Question

At 1.01 atm and 15.0°C a constant - pressure tank has 255 `m^3` of propane. What is the volume of the gas at 48.0°C?

Answer 1

The volume at `"321.2 K"` `("48.0"^"o""C")` is `"251 m"^3"`.

Explanation:

This is an example of the combined gas law, with the equation `(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`.

Given/Known
`P_1="1.01 atm"`
`V_1="225 m"^3"`
`T_1="15.0"^"o""C"+273.15="288.2 K"`
`P_2="1.01 atm"`
`T_2="48.0"^"o""C"+273.15="321.2 K"`

Unknown
`V_2`

Solution
Rearrange the equation to isolate `V_2` and solve.

`V_2=(P_1V_1T_2)/(T_1P_2)`

`V_2=(1.01cancel"atm"xx225"m"^3xx321.2cancel"K")/(288.2cancel"K"xx1.01cancel"atm")="251 m"^3"`