Question

The pressure of a sample of argon at 20°C is decreased from 720 mmHg to 360 mmHg. What was the final temperature of the argon gas?

Answer 1

`T_2 = 146.5 K`

or

`T_2 = -126.5^oC`

Explanation:

For this solution we will use the Gay-Lussac Law relating pressure and temperature in gases. `P_1/T_1 = P_2/T_2`

For this case the values are

`P_1 = 720 mmHg`
`T_1 = 20^oC or 293K`
`P_2 = 360 mmHg`
`T_1 = ?K`

`P_1/T_1 = P_2/T_2`

`(720mmHg)/(293K) = (360mmHg)/T_2`

We can invert both sides to put the missing variable in the numerator.

`(293K)/(720mmHg) = T_2/(360mmHg)`

`(293K)/(720cancel(mmHg))(360cancel(mmHg)) = T_2/(cancel(360mmHg))(cancel(360mmHg))`

`T_2 = 146.5 K`

or

`T_2 = -126.5^oC`