Monday, December 17, 2007

gas laws problems

http://misterguch.brinkster.net/gaslawworksheets.html

148 comments:

ruelyn said...

charr kay maam bay!!!

ruelyn said...

hi maam!!!

Balve said...

Teacher Leni, please post here your designed "On Line Treasure Hunt" and webquest also.

John 07-08 said...

7) Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental research submarine with a volume of 15,000 liters has an internal pressure of 1.2 atm. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250 atm pushing on it, how big will that bubble be?
P1v1=p2v2
V2=p1v1/p2=v2=(1.2)15000/250
ans.72 L
harlene p. hatalan

mark_jeffrey said...

hi mam..

John 07-08 said...

7) Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental research submarine with a volume of 15,000 liters has an internal pressure of 1.2 atm. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250 atm pushing on it, how big will that bubble be?
P1v1=p2v2
V2=p1v1/p2=v2=(1.2)15000/250
ans.72 L
harlene p. hatalan

John 07-08 said...

Hazel Paclar
Boyles law
8.) V1= 0.05L
V2= ?
P1= 250 atm
P2= 50.0 atm

V2= V1P1
------
P2
V2= (0.05L)(250atm)
---------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

John 07-08 said...

stephanie g.sullano
boyles law
8.)v1=0.05L
V2=?
P1=250 atm
p2=50.0 atm

v2=v1p1
-------
p2
v2=(0.05L) (250 atm)
-----------------
v2=12.5
----
50.0
v2=0.25 L,YES

John 07-08 said...

Priscilla S. Akut
Boyles law
8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

John 07-08 said...

2) A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 0C, what will the volume of the balloon be after he heats it to a temperature of 250 0C?
V2=v1t2/t1=v2=(0.4)523.15/293.15
Ans.0.71 L
Harlene p. hatalan

John 07-08 said...

8) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
P1v1=p2v2
V2=p1v1/p2=v2=(0.05)250/50.0=v=0.25L,yes
Racma Mauna 3-john

John 07-08 said...

Priscilla S.Akut
Charles law
3.)On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be?

V1=250 mL
V2=?
T1=19+273.15=292.15
T2=60+273.15=333.15

V2=V1T2
----
T1
V2=333.15/292.15(250 mL)
V2=(1.14)(250 mL)
V2=285 mL

John 07-08 said...

Hazel Paclar
3rd yr John


“ Combined Gas Law “

Problem # 1.


P1 = 12 atm

V1 = 23 L

T1 = 200 K

P2 = 14 atm

T2 = 300 K

V2 = ?

V2 = P1V1T2

P2T1

V2 = 12 atm 300 K ( 23 L )

14 atm 200 K

V2 = ( 0.86) ( 1.5 ) ( 23 L )

V2 = ( 1.29) ( 23 L )

V2 = 29.67 / 29.6

John 07-08 said...

Hazel Paclar
3rd Year – John

CHARLE’S LAW

PROBLEM NO. 2


v1 = 0.4 L
T1 = 20 degrees C + 273.15 = 293.15 K
V2 = ?
T2 = 250 degrees C + 273.15 = 523.15 K

V2 = V1T2

T1


V2 = 523.15 K
293.15 K 0.4 L



V2 = 1.7845

V2 = 0.71 L

John 07-08 said...

stephanie g.sullano
charle's law
PROBLEM NO. 2


v1 = 0.4 L
T1 = 20 degrees C + 273.15 = 293.15 K
V2 = ?
T2 = 250 degrees C + 273.15 = 523.15 K

V2 = V1T2

T1


V2 = 523.15 K
293.15 K 0.4 L



V2 = 1.7845

V2 = 0.71 L

John 07-08 said...

stephanie g.sullano
charle's law
PROBLEM NO. 2


v1 = 0.4 L
T1 = 20 degrees C + 273.15 = 293.15 K
V2 = ?
T2 = 250 degrees C + 273.15 = 523.15 K

V2 = V1T2

T1


V2 = 523.15 K
293.15 K 0.4 L



V2 = 1.7845

V2 = 0.71 L

mark_jeffrey said...

mark jeffrey magparoc
3-john

4) The highest pressure ever produced in a laboratory setting was about 2.0 x 106 atm. If we have a 1.0 x 10-5 liter sample of a gas at that pressure, then release the pressure until it is equal to 0.275 atm, what would the new volume of that gas be? 72.7 L

p1=2000000atm
v1=0.000010 L
p2=0.275 atm
v2=??

p1v1=-p2v2

2000000(0.000010)=0.275(v2)

p1v1/p2=v2

2000000(0.000010)/0.275=v2

20/0.275=v2

v2=72.72727

ANS:
v2=72.7 L

John 07-08 said...

stephanie g. sullano
combined gas law
1.)p1=12 atm
p2=14 atm
v1=23 L
v2=?
T1=200k
T2=300k

p1v1t2
--------
p2t1
v2=(12 atm/14 atm) (300K/200K)(23L)
V2=(0.86) (1.5) (23l)
v2=(1.29)(23 L)
V2=29.67/29.6

John 07-08 said...

Priscilla S.Akut
Combined gas law
1.)If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

P1=12atm
P2=14atm
V1=23L
V2=?
T1=200K
T2=300K

P1V1T2
V2=------
P2T1
V2=(12atm/14atm)(300K/200k)(23L)
V2=(0.86)(1.5)(23L)
V2=(1.29)(23L)
V2=29.6L

John 07-08 said...

2) A gas takes up a volume of 17 liters, has a pressure of 2.3 atm, and a temperature of 299 K. If I raise the temperature to 350 K and lower the pressure to 1.5 atm, what is the new volume of the gas?
P1v1/t1=p2v2/t2
V2=p1v1t2/t1p2
=(2.3)(17 L)(350 K)/(1.5)(299 K)
Ans.30.5
Harlene P. hatalan

John 07-08 said...

Hazel Paclar
1.)Combined gas law
1.)If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

P1=12atm
P2=14atm
V1=23L
V2=?
T1=200K
T2=300K

P1V1T2
V2=------
P2T1
V2=(12atm/14atm)(300K/200k)(23L)
V2=(0.86)(1.5)(23L)
V2=(1.29)(23L)
V2=29.6L

John 07-08 said...

Flower Mae S. Nugas
boyles law
v1=0.05,,v2=?,,p1=250atm,,p2=50.0atm
v2=v1p1/p2
v2=(0.05)(250atm)
50.0atm
v2=12.5
50.0
v2=0.25L,,,yes..

florisa yancha said...

florisa yancha
boyle"s law
1.} P1=1.00 atm


473 mL



T1=473+1.00=474
=474 mL

John 07-08 said...

Flower Mae S. Nugas
combined gas law
p1=12atm,,v1=23L,,T1200k,,P2=14atm,,T2=300k,,V2?,,
V2=P1V1T2/P2T1
V2=(12atm/14atm)(300k/200k)(23L
V2=(0.86)(1.5)(23L)
V2=(1.29)(23L)
V2=29.67/29.6

florisa yancha said...

florisa yancha
charle"s law
1) The temperature inside my refrigerator is about 40 Celsius. If I place a balloon in my fridge that initially has a temperature of 220 C and a volume of 0.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator?
P1=o.5 atm
P2=? atm
4degree Celsius+22 degree Celsius=26
=26

mark_jeffrey said...

Combined Gas Law Problems
mark jeffrey magparoc
3-john

1) If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas? 29.6 L
P1=12 atm
P2=14 atm
V1=23 L
V2=????
T1=200 K
T2=300K

P1v1/t1=p2v2/t2
12 atm(23L)/23L=14 atm(?)/300 K

V2=p1v1t2/t1p2

V2=12 atm(23L)(300K)/200 K(14 atm.)

V2=82800/2800=29.57143

Ans;
29.6 L

florisa yancha said...

florisa yancha
dalton"s law
A metal tank contains three gases: oxygen, helium, and nitrogen. If the partial pressures of the three gases in the tank are 35 atm of O2, 5 atm of N2, and 25 atm of He, what is the total pressure inside of the tank?
65 atm
P1=35 atm,5 atm
P2=25 atm
=65 atm

John 07-08 said...

Flower Mae S. Nugas
charles law no. 2
V1=0.4L,,T1=20 degrees C+273.15=293.15K,,V2=?,,T2=250 degreesC+273.15=523.15K,,V2=V1T1
T1
V2=523.15K
293.15K 0.4L
V2=1.7845
V2=0.71L

gwapoako said...

Charles law
3.)On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be?

V1=250 mL
V2=?
T1=19+273.15=292.15
T2=60+273.15=333.15

V2=V1T2
----
T1
V2=333.15/292.15(250 mL)
V2=(1.14)(250 mL)
V2=285 mL

KARL JULIUS S. DELA CRUZ
III-JOHN

florisa yancha said...

florisa yancha
3-john
the ideal and combined gas laws
3) My car has an internal volume of 2600 liters. If the sun heats my car from a temperature of 200 C to a temperature of 550 C, what will the pressure inside my car be? Assume the pressure was initially 760 mm Hg.

P1=2600 L

P2=?

T1=20 degree Celsius+55
=75

V1=2600 L v2=2600L

mark_jeffrey said...

Charles’ Law Worksheet
mark jeffrey magparoc

3-john

1) The temperature inside my refrigerator is about 40 Celsius. If I place a balloon in my fridge that initially has a temperature of 220 C and a volume of 0.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator? 0.47 L
V1=0.5 L
T2=22’c + 273.15 k=295.15
V2=??
T1=4’c + 273.15k=277.15

V1t1=v2t2
0.5L(277.15)=?(295.15)

V2=0.5(277.15)/295.15

v2=138.5765/295.15

V2=0.469512

Ans:
V2=0.47 L

kheyjaymark said...

4) The highest pressure ever produced in a laboratory setting was about 2.0 x 106 atm. If we have a 1.0 x 10-5 liter sample of a gas at that pressure, then release the pressure until it is equal to 0.275 atm, what would the new volume of that gas be? 72.7 L

p1=2000000atm
v1=0.000010 L
p2=0.275 atm
v2=??

p1v1=-p2v2

2000000(0.000010)=0.275(v2)

p1v1/p2=v2

2000000(0.000010)/0.275=v2

20/0.275=v2

v2=72.72727

ANS:
v2=72.7 L

kheyjaymark B.Longaquit

John 07-08 said...

5) Some students believe that teachers are full of hot air. If I inhale 2.2 liters of gas at a temperature of 180 C and it heats to a temperature of 380 C in my lungs, what is the new volume of the gas?
CHARLES LAW
Racma G. Mauna 3 JOHN
V1=2.2L
V2=?
T1=18+273.15=291.15
T2=38+273.15=311.15
V2=v1t2/t1

V2=311.15/291.15(2.2L)
V2=(1.06)(2.2L)
V2=2.35L

joshua said...

Boyles law
8.) V1= 0.05L
V2= ?
P1= 250 atm
P2= 50.0 atm

V2= V1P1
------
P2
V2= (0.05L)(250atm)
---------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

joshua said...

Charles law
3.)On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be?

V1=250 mL
V2=?
T1=19+273.15=292.15
T2=60+273.15=333.15

V2=V1T2
----
T1
V2=333.15/292.15(250 mL)
V2=(1.14)(250 mL)
V2=285 mL

joshua delvin ryan garces

mary ann said...

boyles law

V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1p1
------
P2
V2=(0.05L)(250atm)
------------
50.0 atm
V2=12.5
-------
50.0
V2+0.25L yes

Mary Ann Hinampas
III-JUDE

aubrey alba said...

aubrey alba

Boyles law
8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

ana marie g. hinampas said...

boyles law

ana marie g. hinampas

V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1p1
------
P2
V2=(0.05L)(250atm)
------------
50.0 atm
V2=12.5
-------
50.0
V2+0.25L yes

ana marie g. hinampas said...

1.)Combined gas law
1.)If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

P1=12atm
P2=14atm
V1=23L
V2=?
T1=200K
T2=300K

P1V1T2
V2=------
P2T1
V2=(12atm/14atm)(300K/200k)(23L)
V2=(0.86)(1.5)(23L)
V2=(1.29)(23L)
V2=29.6L
ana marie g.hinampas

haruka said...
This comment has been removed by the author.
haruka said...

A(n) __________________ is used to lower the energy required to make a reaction take place. It makes chemical reactions go faster without being consumed.

ANSWER: catalyst

ANSWERED BY: alma mae mahasol

haruka said...
This comment has been removed by the author.
almamae said...

The thing we measure when we want to determine the average kinetic energy of random motion in the particles of a substance is _temperature_________________.

ANSWER:temperature

ANSWERED BY:alma mae mahasol

almamae said...
This comment has been removed by the author.
almamae said...
This comment has been removed by the author.
almamae said...

The __________________ is used to describe how much energy is produced or used during a chemical change.

ANSWER:heat of reaction (or enthalpy of reaction)

ANSWERED BY: alma mae mahasol

almamae said...

1) If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature?

ANSWER: 205 K

ANSWERED BY: alma mae mahasol

almamae said...

2) If I have an unknown quantity of gas at a pressure of 1.2 atm, a volume of 31 liters, and a temperature of 87 0C, how many moles of gas do I have?

ANSWER: 1.26 moles

ANSWERED BY: alma mae mahasol

almamae said...

3) If I contain 3 moles of gas in a container with a volume of 60 liters and at a temperature of 400 K, what is the pressure inside the container?

ANSWER:1.64 atm

ANSWERED BY: alma narido

almamae said...

4) If I have 7.7 moles of gas at a pressure of 0.09 atm and at a temperature of 56 0C, what is the volume of the container that the gas is in?

ANSWER:2310 L

ANSWERED BY: alma narido

almamae said...

5) If I have 17 moles of gas at a temperature of 67 0C, and a volume of 88.89 liters, what is the pressure of the gas?

ANSWER:5.34 atm

ANSWERED BY: alma narido

almamae said...

6) If I have an unknown quantity of gas at a pressure of 0.5 atm, a volume of 25 liters, and a temperature of 300 K, how many moles of gas do I have?

ANSWER:0.51 moles

ANSWERED BY: alma narido

almamae said...

7) If I have 21 moles of gas held at a pressure of 78 atm and a temperature of 900 K, what is the volume of the gas?

ANSWER: 19.9 L

ANSWERED BY: alma narido

almamae said...

9) If I have 2.4 moles of gas held at a temperature of 97 0C and in a container with a volume of 45 liters, what is the pressure of the gas?

ANSWER:1.62 atm

ANSWERED BY: raiza kris francisco

almamae said...

10) If I have an unknown quantity of gas held at a temperature of 1195 K in a container with a volume of 25 liters and a pressure of 560 atm, how many moles of gas do I have?

ANSWER:143 moles

ANSWERED BY: raiza kris francisco

almamae said...

11) If I have 0.275 moles of gas at a temperature of 75 K and a pressure of 1.75 atmospheres, what is the volume of the gas?

ANSWER:0.97 L


ANSWERED BY: raiza kris francisco

almamae said...

Another word for freezing is __________________.

ANSWER: fusion
ANSWERED BY: raiza kris francisco

almamae said...

__________________ reactions require energy in order to take place.

ANSWER: endothermic

ANSWERED BY: raiza kris francisco

ana mae manolo said...

The thing we measure when we want to determine the average kinetic energy of random motion in the particles of a substance is_________________. *answer:temperature

ana mae manolo said...

The thing we measure when we want to determine the average kinetic energy of random motion in the particles of a substance is_________________.
*answer:temperature

ana mae manolo said...

1.00 L of a gas at standard temperature and pressure is compressed to 473 mL. What is the new pressure of the gas? *answer;2.11 atm

ana mae manolo said...

Part of the reason that conventional explosives cause so much damage is that their detonation produces a strong shock wave that can knock things down. While using explosives to knock down a building, the shock wave can be so strong that 12 liters of gas will reach a pressure of 3.8 x 104 mm Hg. When the shock wave passes and the gas returns to a pressure of 760 mm Hg, what will the volume of that gas be? *answer:600L

ana mae manolo said...

he temperature inside my refrigerator is about 40 Celsius. If I place a balloon in my fridge that initially has a temperature of 220 C and a volume of 0.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator? *answer:0.47L

ana mae manolo said...

A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 0C, what will the volume of the balloon be after he heats it to a temperature of 250 0C? *answer:0.71L
submitted by CHERIE MAE GALARPE

ana mae manolo said...

Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver? answer:v=0.25L,yes submitted by CHERIE MAE GALARPE

kimjay dulalas said...

Boyles’ Law

1) 1.00 L of a gas at standard temperature and pressure is compressed to 473 mL. What is the new pressure of the gas?
Given:
V1 =1.00L P1=1atm V2=473mL P2=?

Volume is inversely proportion to pressure. A decrease in volume means an increase in pressure. Hence, we multiply P1 by a factor of volume that will increase its value. FORMULA: P1V1=P1V2
SOLUTION:
P2=(1atm)(1000ml)
473mL
P2=2.11atm
Kimjay B.Dulalas

kimjay dulalas said...

Charles’ Law Worksheet

1) The temperature inside my refrigerator is about 40 Celsius. If I place a balloon in my fridge that initially has a temperature of 220 C and a volume of 0.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator?

Given:
V1= 0.5L T1 =40 C + 273 K
V2= ? T2= 220 C + 273 K

Formula:
V1 = V2
T1 T2


Solution:
V2= V1T2
T1

V2= (0.5)(277 K)
295 K
V2 = 0.47 L

)

kimjay dulalas said...

Combined Gas Law Problems



Use the combined gas law to solve the following problems:


1) If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

Given:
V1 = 23 L P1 = 12 atm T1 = 200 K
V2 = ? P2 = 14 atm T2 = 300 K

Formula:
P1V1 = P2V2
T1 T2
Solution:
V2 = P1V1T2
T1P2

V2= (12 atm)(23 L)(300 K)
(200 K)(14 atm)

V2 = 29.6 L

John 07-08 said...

m mercy tuhod boyleslaw 8} V1=0.05L V2? ...P1=250atm P2=50.0atm V2=V1P1 ...V2={0.05L}{250atm} ...50atm V2=12.5 ...50.0 v2=0.25L ...ANS.YES!

John 07-08 said...

mercy tuhod combined gas laws#1 T1=? P1=5.4atm V1=120L T1=P1V1 T1=5.4atmX120L T1=648.

John 07-08 said...

mercy tuhod charles law#1 T1=4CELCIUS T2=22 V1=0.5L V2=? V2=V1XT2/T1 V2=0.5X22/4 V2=2.75

John 07-08 said...

mercytuhod gaslaws#1 P1=5.6atm V1=12L T1? T1=P1V1 T1=5.6atm X12L T1=67.2

gezelle marie said...

gezelle marie japona charle's law 2) A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 0C, what will the volume of the balloon be after he heats it to a temperature of 250 0C?
PROBLEM NO. 2


v1 = 0.4 L
T1 = 20 degrees C + 273.15 = 293.15 K
V2 = ?
T2 = 250 degrees C + 273.15 = 523.15 K

V2 = V1T2

T1


V2 = 523.15 K
293.15 K 0.4 L



V2 = 1.7845

V2 = 0.71

gezelle marie said...

gezelle marie japona combined gas law problem 1) If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

Problem # 1.


P1 = 12 atm

V1 = 23 L

T1 = 200 K

P2 = 14 atm

T2 = 300 K

V2 = ?

V2 = P1V1T2

P2T1

V2 = 12 atm 300 K ( 23 L )

14 atm 200 K

V2 = ( 0.86) ( 1.5 ) ( 23 L )

V2 = ( 1.29) ( 23 L )

V2 = 29.67 / 29.6

gezelle marie said...

gezelle marie japona dalton's law A metal tank contains three gases: oxygen, helium, and nitrogen. If the partial pressures of the three gases in the tank are 35 atm of O2, 5 atm of N2, and 25 atm of He, what is the total pressure inside of the tank?
65 atm
P1=35 atm,5 atm
P2=25 atm
=65 atm

gezelle marie said...

gezelle marie8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L japona

CnAgReLtO said...

Carlo Montano M. Radaza
3-John

V1=0.05L
V2=?

P1=250 atm
P2=50.0 atm

V2=P1V1/V2
--------------
V2=(250)(0.05)/50.0
--------------
V2=12.5/50.0

V2=0.25L,YES

jenifer said...

Ma.Jenifer G. Tibay Boyles Low 8.V1=0.05L V2=? P1=250atm P2=50.0atm v2=v1p1 _______ p1 V2=(0.05L)(250atm) _________________ 50.0atm V2=12.5 _______ 50.0 V2=0.25L , YES

jenifer said...

V1=0.05
V2=?
P1=250 atom
P2=50.0 atom

V2=P1V1/P2

V2=(250)(0.05)/50.0L

V2=12.5/50.0L

V2=0.25L,YES

geamp said...

V1=0.05L
V2=?
P1=250
P2=50.0

V2=V1P1/V2

V2=(0.05)(250)/50.0

V2=12.5/50.0

V2=0.25L,YES

shanenna said...

charles law
1) The temperature inside my refrigeratoris about 4 degrees Celcius.if i place a ballonin my fridge initially has a temp. of 22 degrees celcius and a volume of 0.5 liters.what will be the volume of the balloon when its fully cooled in my refrigerator?




answer:
FORMULA:
V1/v2=t1/t2

v1= 0.5
V2=?
t1=4 degrees celcius
t2=22 degrees celcius
t1=4+273k= 277k
t2=22+273k=295k
(277k)(0.5)/ 295k

answer:0.46949153
round of to 0.47

shanenna said...

Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?


answer:
8.) V1= 0.05L
V2= ?
P1= 250 atm
P2= 50.0 atm

V2= V1P1
------
P2
V2= (0.05L)(250atm)
---------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

ruelyn said...

ruelyn yecyec

Boyles law
8.) V1= 0.05L
V2= ?
P1= 250 atm
P2= 50.0 atm

V2= V1P1
------
P2
V2= (0.05L)(250atm)
---------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

ruelyn said...

ruelyn yecyec
problem no.2

1) If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

Given:
V1 = 23 L P1 = 12 atm T1 = 200 K
V2 = ? P2 = 14 atm T2 = 300 K

Formula:
P1V1 = P2V2
T1 T2
Solution:
V2 = P1V1T2
T1P2

V2= (12 atm)(23 L)(300 K)
(200 K)(14 atm)

V2 = 29.6 L

ruelyn said...

ruelyn yecyec
Boyles’ Law

1) 1.00 L of a gas at standard temperature and pressure is compressed to 473 mL. What is the new pressure of the gas?
Given:
V1 =1.00L P1=1atm V2=473mL P2=?

Volume is inversely proportion to pressure. A decrease in volume means an increase in pressure. Hence, we multiply P1 by a factor of volume that will increase its value. FORMULA: P1V1=P1V2
SOLUTION:
P2=(1atm)(1000ml)
473mL
P2=2.11atm

rosemarie said...

rosemarie kiamco

Boyles law
8.) V1= 0.05L
V2= ?
P1= 250 atm
P2= 50.0 atm

V2= V1P1
------
P2
V2= (0.05L)(250atm)
---------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

rosemarie said...

rosemarie kiamco

Charles law
3.)On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be?

V1=250 mL
V2=?
T1=19+273.15=292.15
T2=60+273.15=333.15

V2=V1T2
----
T1
V2=333.15/292.15(250 mL)
V2=(1.14)(250 mL)
V2=285 mL

rosemarie said...

rosemarie kiamco

4) The highest pressure ever produced in a laboratory setting was about 2.0 x 106 atm. If we have a 1.0 x 10-5 liter sample of a gas at that pressure, then release the pressure until it is equal to 0.275 atm, what would the new volume of that gas be? 72.7 L

p1=2000000atm
v1=0.000010 L
p2=0.275 atm
v2=??

p1v1=-p2v2

2000000(0.000010)=0.275(v2)

p1v1/p2=v2

2000000(0.000010)/0.275=v2

20/0.275=v2

v2=72.72727

ANS:
v2=72.7 L

rachelle bahan said...

rachelle bahancharle's law 2) A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 0C, what will the volume of the balloon be after he heats it to a temperature of 250 0C?
PROBLEM NO. 2


v1 = 0.4 L
T1 = 20 degrees C + 273.15 = 293.15 K
V2 = ?
T2 = 250 degrees C + 273.15 = 523.15 K

V2 = V1T2

T1


V2 = 523.15 K
293.15 K 0.4 L



V2 = 1.7845

V2 = 0.71

rachelle bahan said...

8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L

rachelle bahan said...

dalton's law A metal tank contains three gases: oxygen, helium, and nitrogen. If the partial pressures of the three gases in the tank are 35 atm of O2, 5 atm of N2, and 25 atm of He, what is the total pressure inside of the tank?
65 atm
P1=35 atm,5 atm
P2=25 atm
=65 atm

rachelle bahan said...

combined gas law problem 1) If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

Problem # 1.


P1 = 12 atm

V1 = 23 L

T1 = 200 K

P2 = 14 atm

T2 = 300 K

V2 = ?

V2 = P1V1T2

P2T1

V2 = 12 atm 300 K ( 23 L )

14 atm 200 K

V2 = ( 0.86) ( 1.5 ) ( 23 L )

V2 = ( 1.29) ( 23 L )

V2 = 29.67 / 29.6

mae_92@yahoo.com said...

Erlin Mae Chan Boyles Law 8.)V1=0.05L V2=? P1=250atm P2=50.0atm V2=V1P1 ----P2 V2=(0.05L)(250atm) ----------- 50.0atm V=12.5 ----- 50.0 V2=0.25L,yes

mae_92@yahoo.com said...

Erlin Mae Chan Combined Gas Law 1.)P1=12atm V1=23L T1=200K P2=14atm T2=300K V2=? V2=P1V1P2 ----- P2T1 V2=12atm 300k(23L) ------ 14atm 200k V2=(0.86)(1.5)(23L) V2=(1.29)(23L) V2=29.6

jonalyn said...

jonalyn hernaiz
Boyles law
8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

March 11,2008 1:30PM III-JUDE

jonalyn said...

maurice sagbigsal

boyles law

V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1p1
------
P2
V2=(0.05L)(250atm)
------------
50.0 atm
V2=12.5
-------
50.0
V2+0.25L yes

III-JUDE maurice sagbigsal

jonalyn said...

maricho macas

Combined gas law

1.)If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

P1=12atm
P2=14atm
V1=23L
V2=?
T1=200K
T2=300K

P1V1T2
V2=------
P2T1
V2=(12atm/14atm)(300K/200k)(23L)
V2=(0.86)(1.5)(23L)
V2=(1.29)(23L)
V2=29.6L

III-JUDE maricho macas

jonalyn said...

charity raluto

boyles law
v1=0.05,,v2=?,,p1=250atm,,p2=50.0atm
v2=v1p1/p2
v2=(0.05)(250atm)
50.0atm
v2=12.5
50.0
v2=0.25L,,,yes..

III-JUDE charity raluto

rosemarie said...

rosemarie dacuyan

III-jude

boyles law
8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

March 12,2008

rosemarie dacuyan

jesivel said...

jesivel jadap combined gas law problem 1) If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

Problem # 1.


P1 = 12 atm

V1 = 23 L

T1 = 200 K

P2 = 14 atm

T2 = 300 K

V2 = ?

V2 = P1V1T2

P2T1

V2 = 12 atm 300 K ( 23 L )

14 atm 200 K

V2 = ( 0.86) ( 1.5 ) ( 23 L )

V2 = ( 1.29) ( 23 L )

V2 = 29.67 / 29.6
march 12 200

jesivel jadap
III-JUDE

jesivel said...

jesivel jadap
III-JUDE
3.)On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be?

V1=250 mL
V2=?
T1=19+273.15=292.15
T2=60+273.15=333.15

V2=V1T2
----
T1
V2=333.15/292.15(250 mL)
V2=(1.14)(250 mL)
V2=285 mL

JESIVEL JADAP
III-JUDE

Danica Rose D. Arsenal said...

Combined gas law
1.)If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

P1=12atm
P2=14atm
V1=23L
V2=?
T1=200K
T2=300K

P1V1T2
V2=------
P2T1
V2=(12atm/14atm)(300K/200k)(23L)
V2=(0.86)(1.5)(23L)
V2=(1.29)(23L)
V2=29.6L

January 10, 2008 3:58 PM

noemi_fabria@yahoo.com said...

noemi fabria said...
boyles law

V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1p1
------
P2
V2=(0.05L)(250atm)
------------
50.0 atm
V2=12.5
-------
50.0
V2+0.25L yes

lorr said...

lorraine g.calam
III-JUDE

7) Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental research submarine with a volume of 15,000 liters has an internal pressure of 1.2 atm. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250 atm pushing on it, how big will that bubble be?
P1v1=p2v2
V2=p1v1/p2=v2=(1.2)15000/250
ans.72 L
LORRAINE G.CALAM

MARCH 12,2008

borjal_marianne@yahoo.com said...

ma.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25l yes marianne borjal 11

karen_ganinay@yahoo.comj said...

KaRen sAiD...
boyles law
v1=0.05,,v2=?,,p1=250atm,,p2=50.0atm
v2=v1p1/p2
v2=(0.05)(250atm)
50.0atm
v2=12.5
50.0
v2=0.25L,,,yes..

jesivel said...

angel rose marie ybanez combined gas law problem 1) If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

Problem # 1.


P1 = 12 atm

V1 = 23 L

T1 = 200 K

P2 = 14 atm

T2 = 300 K

V2 = ?

V2 = P1V1T2

P2T1

V2 = 12 atm 300 K ( 23 L )

14 atm 200 K

V2 = ( 0.86) ( 1.5 ) ( 23 L )

V2 = ( 1.29) ( 23 L )

V2 = 29.67 / 29.6
march 12 200

angel rose marie ybanez
III-JUDE

March 11, 2008 9:28 PM

karen_ganinay@yahoo.comj said...

combined gas law

Conson,Kevin Jon C.
III-Jude

p1=12atm,,v1=23L,,T1200k,,P2=14atm,,T2=300k,,V2?,,
V2=P1V1T2/P2T1
V2=(12atm/14atm)(300k/200k)(23L
V2=(0.86)(1.5)(23L)
V2=(1.29)(23L)
V2=29.67/29.6

Boyles law
8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

rapog said...

rapog sheila
III-JUDE

Boyles law
8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

MARCH 12,2OO8

karen_ganinay@yahoo.comj said...

8) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
P1v1=p2v2
V2=p1v1/p2=v2=(0.05)250/50.0=v=0.25L,yes KaReN GANINAY III-JUDE....

jesivel said...

boyles law
angel rose marie ybanez
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1p1
------
P2
V2=(0.05L)(250atm)
------------
50.0 atm
V2=12.5
-------
50.0
V2+0.25L yes

March 11, 2008 9:33 PM

michelle_obeys@yahoo.com said...

Combined gas law
1.)If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

P1=12atm
P2=14atm
V1=23L
V2=?
T1=200K
T2=300K

P1V1T2
V2=------
P2T1
V2=(12atm/14atm)(300K/200k)(23L)
V2=(0.86)(1.5)(23L)
V2=(1.29)(23L)
V2=29.6L
ana marie g.hinampas

January 11, 2008 3:20 AM


Comment deleted

jesivel said...

klyndon Boyles law
8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes
genson III-JUDE

kevin jon conson said...

Charles law
3.)On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be?

V1=250 mL
V2=?
T1=19+273.15=292.15
T2=60+273.15=333.15

V2=V1T2
----
T1
V2=333.15/292.15(250 mL)
V2=(1.14)(250 mL)
V2=285 mL

kevin jon conson
III-jude

michelle_obeys@yahoo.com said...

Combined gas law
1.)If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

P1=12atm
P2=14atm
V1=23L
V2=?
T1=200K
T2=300K

P1V1T2
V2=------
P2T1
V2=(12atm/14atm)(300K/200k)(23L)
V2=(0.86)(1.5)(23L)
V2=(1.29)(23L)
V2=29.6L
ana marie g.hinampas

January 11, 2008 3:20 AM


Comment deleted

analeth said...

analeth cartajena
III-JUDE.)On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be?

V1=250 mL
V2=?
T1=19+273.15=292.15
T2=60+273.15=333.15

V2=V1T2
----
T1
V2=333.15/292.15(250 mL)
V2=(1.14)(250 mL)
V2=285 mL

analeth cartajena
III-JUDE

jesivel said...

jesivel jadap IBoyles law
8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

II-JUDE

klyndon said...

klyndon genson
III-JUDE

7) Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental research submarine with a volume of 15,000 liters has an internal pressure of 1.2 atm. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250 atm pushing on it, how big will that bubble be?
P1v1=p2v2
V2=p1v1/p2=v2=(1.2)15000/250
ans.72 L
harlene p. hatalan

vernedette said...

vernedette dagaya
III-JUDE

Charles’ Law Worksheet

1) The temperature inside my refrigerator is about 40 Celsius. If I place a balloon in my fridge that initially has a temperature of 220 C and a volume of 0.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator?

Given:
V1= 0.5L T1 =40 C + 273 K
V2= ? T2= 220 C + 273 K

Formula:
V1 = V2
T1 T2


Solution:
V2= V1T2
T1

V2= (0.5)(277 K)
295 K
V2 = 0.47 L

E

III-JUDE

vernedette said...

vernedette dagaya
III-JUDE

Charles’ Law Worksheet

1) The temperature inside my refrigerator is about 40 Celsius. If I place a balloon in my fridge that initially has a temperature of 220 C and a volume of 0.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator?

Given:
V1= 0.5L T1 =40 C + 273 K
V2= ? T2= 220 C + 273 K

Formula:
V1 = V2
T1 T2


Solution:
V2= V1T2
T1

V2= (0.5)(277 K)
295 K
V2 = 0.47 L

E

III-JUDE

vernedette said...

jonalyn hernaiz

III-JUDEcharle's law
PROBLEM NO. 2


v1 = 0.4 L
T1 = 20 degrees C + 273.15 = 293.15 K
V2 = ?
T2 = 250 degrees C + 273.15 = 523.15 K

V2 = V1T2

T1


V2 = 523.15 K
293.15 K 0.4 L



V2 = 1.7845

V2 = 0.71 L

III-JUDE
JOnalyn CONcepcion Romea Hernaiz

angel_rose_ybanez@yahoo.com said...

AnGel RoSe YBANez
III-jUdE..

2) A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 0C, what will the volume of the balloon be after he heats it to a temperature of 250 0C?
V2=v1t2/t1=v2=(0.4)523.15/293.15
Ans.0.71 L

lord neil actub said...

lord neil C. actub
lll-jude

Charles law
3.)On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be?

V1=250 mL
V2=?
T1=19+273.15=292.15
T2=60+273.15=333.15

V2=V1T2
----
T1
V2=333.15/292.15(250 mL)
V2=(1.14)(250 mL)
V2=285 mL

angel_rose_ybanez@yahoo.com said...

Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?


answer:
8.) V1= 0.05L
V2= ?
P1= 250 atm
P2= 50.0 atm

V2= V1P1
------
P2
V2= (0.05L)(250atm)
---------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

angel_rose_ybanez@yahoo.com said...

Combined gas law
1.)If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas?

P1=12atm
P2=14atm
V1=23L
V2=?
T1=200K
T2=300K

P1V1T2
V2=------
P2T1
V2=(12atm/14atm)(300K/200k)(23L)
V2=(0.86)(1.5)(23L)
V2=(1.29)(23L)
V2=29.6L

karen_ganinay@yahoo.comj said...

1.00 L of a gas at standard temperature and pressure is compressed to 473 mL. What is the new pressure of the gas?
Given:
V1 =1.00L P1=1atm V2=473mL P2=?

Volume is inversely proportion to pressure. A decrease in volume means an increase in pressure. Hence, we multiply P1 by a factor of volume that will increase its value. FORMULA: P1V1=P1V2
SOLUTION:
P2=(1atm)(1000ml)
473mL
P2=2.11atm

gerald cabardo said...

Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental research submarine with a volume of 15,000 liters has an internal pressure of 1.2 atm. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250 atm pushing on it, how big will that bubble be?
P1v1=p2v2
V2=p1v1/p2=v2=(1.2)15000/250
ans.72 L

gerald cabardo said...

Boyles law
8.) V1= 0.05L
V2= ?
P1= 250 atm
P2= 50.0 atm

V2= V1P1
------
P2
V2= (0.05L)(250atm)
---------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

sandramae said...

SANDRA MAE CAIREL
III-JUDE


Charles law
3.)On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be?

V1=250 mL
V2=?
T1=19+273.15=292.15
T2=60+273.15=333.15

V2=V1T2
----
T1
V2=333.15/292.15(250 mL)
V2=(1.14)(250 mL)
V2=285 mL

SANDRA MAE CAIREL
MARCH 12,2008

sandramae said...

SANDRA MAE CAIREL
III-JUDE


Charles law
3.)On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be?

V1=250 mL
V2=?
T1=19+273.15=292.15
T2=60+273.15=333.15

V2=V1T2
----
T1
V2=333.15/292.15(250 mL)
V2=(1.14)(250 mL)
V2=285 mL

SANDRA MAE CAIREL
MARCH 12,2008

dolly said...

DOLLY GRACE ACERA
III-JUDE


8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L

MARCH 12,2008

dolly said...

DOLLY GRACE ACERA
III-JUDE


8.) Divers get “the bends” if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.05 L of gas in his blood under a pressure of 250 atm, then rises instantaneously to a depth where his blood has a pressure of 50.0 atm, what will the volume of gas in his blood be? Do you think this will harm the diver?
V1=0.05L
V2=?
P1=250 atm
P2=50.0 atm

V2=V1P1
----
P2
V2=(0.05L)(250 atm)
----------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L

MARCH 12,2008

kevin jon conson said...

7) Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental research submarine with a volume of 15,000 liters has an internal pressure of 1.2 atm. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250 atm pushing on it, how big will that bubble be?
P1v1=p2v2
V2=p1v1/p2=v2=(1.2)15000/250
ans.72 L

kevin jon conson said...

kevin jon conson
III-jude

Boyles law
8.) V1= 0.05L
V2= ?
P1= 250 atm
P2= 50.0 atm

V2= V1P1
------
P2
V2= (0.05L)(250atm)
---------------
50.0 atm
V2= 12.5
----
50.0
V2= 0.25L yes

kevin jon conson said...

Charles law
3.)On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be?

V1=250 mL
V2=?
T1=19+273.15=292.15
T2=60+273.15=333.15

V2=V1T2
----
T1
V2=333.15/292.15(250 mL)
V2=(1.14)(250 mL)
V2=285 mL

kevin jon conson said...

kevin jon conson
III-jude

Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on them. An experimental research submarine with a volume of 15,000 liters has an internal pressure of 1.2 atm. If the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250 atm pushing on it, how big will that bubble be?
P1v1=p2v2
V2=p1v1/p2=v2=(1.2)15000/250
ans.72 L

kevin jon conson said...

2) A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 0C, what will the volume of the balloon be after he heats it to a temperature of 250 0C?
V2=v1t2/t1=v2=(0.4)523.15/293.15
Ans.0.71 L

march 12,2008

kevin jon conson said...

CHARLE’S LAW

PROBLEM NO. 2


v1 = 0.4 L
T1 = 20 degrees C + 273.15 = 293.15 K
V2 = ?
T2 = 250 degrees C + 273.15 = 523.15 K

V2 = V1T2

T1


V2 = 523.15 K
293.15 K 0.4 L



V2 = 1.7845

V2 = 0.71 L

March 12, 2008

kevin jon conson said...

4) The highest pressure ever produced in a laboratory setting was about 2.0 x 106 atm. If we have a 1.0 x 10-5 liter sample of a gas at that pressure, then release the pressure until it is equal to 0.275 atm, what would the new volume of that gas be? 72.7 L

p1=2000000atm
v1=0.000010 L
p2=0.275 atm
v2=??

p1v1=-p2v2

2000000(0.000010)=0.275(v2)

p1v1/p2=v2

2000000(0.000010)/0.275=v2

20/0.275=v2

v2=72.72727

ANS:
v2=72.7 L

March 12, 2008

kevin jon conson said...

combined gas law
1.)p1=12 atm
p2=14 atm
v1=23 L
v2=?
T1=200k
T2=300k

p1v1t2
--------
p2t1
v2=(12 atm/14 atm) (300K/200K)(23L)
V2=(0.86) (1.5) (23l)
v2=(1.29)(23 L)
V2=29.67/29.6

March 12, 2008

kevin jon conson said...

dalton"s law
A metal tank contains three gases: oxygen, helium, and nitrogen. If the partial pressures of the three gases in the tank are 35 atm of O2, 5 atm of N2, and 25 atm of He, what is the total pressure inside of the tank?
65 atm
P1=35 atm,5 atm
P2=25 atm
=65 atm

March 12, 2008

kevin jon conson said...

5) Some students believe that teachers are full of hot air. If I inhale 2.2 liters of gas at a temperature of 180 C and it heats to a temperature of 380 C in my lungs, what is the new volume of the gas?
CHARLES LAW
Racma G. Mauna 3 JOHN
V1=2.2L
V2=?
T1=18+273.15=291.15
T2=38+273.15=311.15
V2=v1t2/t1

V2=311.15/291.15(2.2L)
V2=(1.06)(2.2L)
V2=2.35L

March 12, 2008

kevin jon conson said...

6) If I have an unknown quantity of gas at a pressure of 0.5 atm, a volume of 25 liters, and a temperature of 300 K, how many moles of gas do I have?

ANSWER:0.51 moles

March 12, 2008

kevin jon conson said...

9) If I have 2.4 moles of gas held at a temperature of 97 0C and in a container with a volume of 45 liters, what is the pressure of the gas?

ANSWER:1.62 atm

March 12, 2008

kevin jon conson said...

10) If I have an unknown quantity of gas held at a temperature of 1195 K in a container with a volume of 25 liters and a pressure of 560 atm, how many moles of gas do I have?

ANSWER:143 moles

March 12, 2008

kevin jon conson said...

11) If I have 0.275 moles of gas at a temperature of 75 K and a pressure of 1.75 atmospheres, what is the volume of the gas?

ANSWER:0.97 L

March 12, 2008

angel_rose_ybanez@yahoo.com said...

hi maam!!!!!!!!!!!!!!!!!!!!!!!!

cutie..shiedden said...
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