Chemistry I-Honors
Gas Laws Problem Set #1
Answer Key
1.
2.
72.3 cm Hg = 723 torr =
96.4 kPa = 0.951 atm. = 14.0 psi
133 psi = 6880 mm Hg =
o
3. 25.0 C = 298 K
917 kPa =
271 in Hg =
o
-130.0 C = 143 K
9.05 atm.
o
50.0 K = -223 C
o
4. At constant volume, if one mole of a gas at 800.0 mm Hg pressure and 300 C has the pressure increased to
o
1200.0 mm Hg, what will be the new temperature (in C)? P1 / T1 = P2 / T2
T2 = P2 x T1 / P1 = (1200.0 mm)(573 K) / 800.0 mm =
859.5 K =
587oC
5. Consider a set amount of helium gas in a fixed pressure situation (normal lab conditions are fixed pressure).
If this gas is originally contained in a cylinder with a movable piston and the initial volume is 450.0 ml at
2.00 atm, what will be the new volume if the pressure drops to 0.755 atm?
V1 x P1 = V2 x P2
V2 =
[ 450.0 ml x 2.00 atm ] / 0.755 atm = 1192 ml
=
1190 ml
6. What kind of pressure change is required to condense 12.0 liters of a gas into a 2.00-liter container?
V1 x P1 = V2 x P2
==>
P2 = (12.0 liters x P1) / 2.00 liters = 6 x P1
need to increase the pressure six times the original amount
7. A gas is put into a 3.00-liter container with a movable piston at a pressure of 745 mm Hg and a temperature
o
o
of 25.0 C. If the temperature is increased to 100.0 C and the pressure increases to 900.0 mm Hg, what will
be the new volume? V1 P1 / T1 = V2P2 / T2
V2
= [(3.00 liter)(745 mm)(373 K)] / [(298 K)(900.0 mm)] = 3.11 liters
8. A gas originally starts out at STP in a 5.00 liter vessel. The pressure is then tripled. If the container is
rigid, what will be the new temperature (in Kelvin) of the gas? V1 P1 / T1 = V2P2 / T2 & V1 = V2
T2
= [(3.00 atm)273 K)] / (1.00 atm) = 819 K