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
Ernest Z.
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).

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