Question

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

Answer 1

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).