# 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?

Dec 9, 2015

The volume at $\text{321.2 K}$ $\left(\text{48.0"^"o""C}\right)$ is $\text{251 m"^3}$.

#### Explanation:

This is an example of the combined gas law, with the equation $\frac{{P}_{1} {V}_{1}}{{T}_{1}} = \frac{{P}_{2} {V}_{2}}{{T}_{2}}$.

Given/Known
${P}_{1} = \text{1.01 atm}$
${V}_{1} = \text{225 m"^3}$
${T}_{1} = \text{15.0"^"o""C"+273.15="288.2 K}$
${P}_{2} = \text{1.01 atm}$
${T}_{2} = \text{48.0"^"o""C"+273.15="321.2 K}$

Unknown
${V}_{2}$

Solution
Rearrange the equation to isolate ${V}_{2}$ and solve.

${V}_{2} = \frac{{P}_{1} {V}_{1} {T}_{2}}{{T}_{1} {P}_{2}}$

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