# Gas Law Problems #1 CV

```LA CEIBA BILINGUAL SCHOOL
PHYSICAL SCIENCE STUDY GUIDE
GAS LAW PROBLEMS #1
Name: Cesar Vivas Grade 10th _____#List: ______ Date: _____
GAS LAW PROBLEMS
1 atm = 101.3 kPa = 760 mm Hg; K = 273 + oC; 1 L = 1,000 mL; R = 0.0821 (atm.L)/(mol.R); 8.31 (Kpa.L)/(mol.R)
Boyle’s Law Problems:
P1VI=P2V2
1.If 22.5 L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at constant temperature. What is the
new volume?
 P1V1=P2V2
 V2=P1V1P2
 V2=748⋅22/5725
 V2=23.2L
2.A gas with a volume of 4.0L at a pressure of 205kPa is allowed to expand to a volume of 12.0L. What
is the pressure in the container if the temperature remains constant?
 P2=P1V1V2
 P2=205kPa&times;4.012.0
 P2=68 kPa
3.What pressure is required to compress 196.0 liters of air at 1.00 atmosphere into a cylinder whose
volume is 26.0 liters?
 P2=P1xV1/V2
 P2 = 196 atm / 26
 P2=7.53 atm
4.A 40.0 L tank of ammonia has a pressure of 12.7 kPa. Calculate the volume of the ammonia if its
pressure is changed to 8.4 kPa while its temperature remains constant.
V2 = P1V1 / P2

 V2 = 12.7 kPa x 40.0 L / 8.4 kPa
 V2= 60.5 L
LA CEIBA BILINGUAL SCHOOL
PHYSICAL SCIENCE STUDY GUIDE
GAS LAW PROBLEMS #1
Name: Cesar Vivas Grade 10th _____#List: ______ Date: _____
Charles’ Law Problems:
V1T2=V2T1
1.Calculate the decrease in temperature when 6.00 L at 20.0 &deg;C is compressed to 4.00 L.




6.00 X 293 = 4.00 X T2
1758 = 4.00xT2
1758/4.00 =T2
439.5 = T2
2.A container containing 5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What
must the new temperature be in order to maintain the same pressure (as required by Charles' Law)?
 T=T1xV2/V1
 2000 / 5.00=T2
 T=400K
3.A gas occupies 900.0 mL at a temperature of 27.0 &deg;C. What is the volume at 132.0 &deg;C?
 V2=V1T1&times;T2
 V2=900.0mL/300K&times;392K
 V2=1176 K
4.If 15.0 liters of neon at 25.0 &deg;C is allowed to expand to 45.0 liters, what must the new temperature be
to maintain constant pressure?
 15.0 L / 298.0 K = 45.0 L / T2
 298 x 45 / 15 = T2
 T2=894 K
```