Question

A sample of argon with a volume of 6.18 L, a pressure of 761 torr, and a temperature of 20.0 degree C expended to a volume of 9.45 L and a pressure of 373 torr. What was its final temperature in degree C?

Answer 1

TThe final temperature will be `"-53.4"^("o")"C"`.

You will need to use the combined gas law in order to solve this problem. The equation is:

`(P_1V_1)/(T_1) = (P_2V_2)/(T_2)`

Known/Given:
`P_1 = 761 "torr"`
`V_1 = 6.18 L`
`T_1 = 20.0"^(o)C + 273.15 = 293.2 K` (temp must be in Kelvins)
`P_2 = 373 "torr"`
`V_2 = 9.45 L`

Unknown:
`T_2` in degrees Celsius

Solution: Rearrange the combined gas law equation to isolate `T"2`. Then solve for `T_2`.

`T_2 = (P_2V_2T_1)/(P_1V_1) = (373 cancel ("torr")*9.45 cancel ( L)*293.2 K)/(761 cancel ("torr")*6.18 cancel (L)) = 219.8 K`

Convert temperature from Kelvins to degrees Celsius.

`219.8 K - 273.15 = -53.4"^(o)C`