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

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 × $\left(303.2 \text{ K")/(293.2" K}\right)$ = 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).