Question

The volume of an ideal gas is increased from 0.61 m3 to 3.9 m3 while maintaining a constant pressure of 970 Pa (1 Pa = 1 N/m2). If the initial temperature of the gas is 65 K, (a) What is the final temperature of the gas?

Answer 1

416K

Explanation:

Given:

Initial volume
`V_(1)`= 0.61 m^3 => 610L

Final volume
`V_(2)`= 3.9 m^3 => 3900L

Initial temperature
`T_(1)` = 65 K

Final temperature
`T_(2)`=?

Charles law states that the volume of an ideal gas at constant pressure is directly proportional to the absolute temperature.

Therefore,

`V_(1)` /
`T_(1)` =
`V_(2)` /
`T_(2)`

610/ 65 = 3900 /
`T_(2)`

`T_(2)` = (3900 x 65) / 610

`T_(2)`= 415.5 K ≈ 416K

The final temperature of the gas is 416K

Answer 2

Final Temperature of the gas = 416K

Explanation:

Using the Charles Law;

V₁/T₁ = V₂/T₂

V₁ = 0.61m³, V₂=3.9m³, T₁= 65K, T₂ = ?

T₂ = V₂T₁/V₁ = (3.9 x 65)/0.61 = 415.57377K ≅ 416K