Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 ˚C to 30.0 ˚C.

1 Answer
May 8, 2014

The pressure increases by 0.03 atm.

This problem involves Gay-Lussac's Law. It states that the pressure exerted on the sides of a container by an ideal gas of fixed volume is proportional to its temperature.

#P_1/T_1=P_2/T_2#

#P_1# = 1.00 atm; #T_1# = (20.0 + 273.15) K = 293.2 K.
#P_2# = ?; #T_2# = (30.0 + 273.15) K = 303.2 K.

We know #P_1#, #T_1#, and #T_2#. Thus, we can calculate #P_2#.

#P_2 = P_1 × T_2/T_1# = 1.00 atm × #(303.2" K")/(293.2" K")# = 1.03 atm

#ΔP = P_2 – P_1# = (1.03 -1.00) atm = 0.03 atm

The pressure increases by 0.03 atm.

This makes sense. The temperature increases by 10 parts in 300 or 3 parts in 100 (3 %). So the pressure should increase by about 3 % (0.03 atm).