# A gas takes up a volume of 19.5 liters, has a pressure of 2.9 atm, and a temperature of 5.5°C. If I raise the temperature to 65°C and lower the pressure to 1.5 atm, what is the new volume of the gas?

Feb 1, 2016

${V}_{2} = 45.8 L$

#### Explanation:

From the ideal gas law $P V = n R T$ we can conclude that $n = \frac{P V}{R T}$.

If the pressure $P$, the volume $V$ and the temperature $T$ of the gas change between two points, this change can be illustrated by:

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

Therefore, this expression can be modified as:

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

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

${V}_{2} = \frac{2.9 \cancel{a t m} \times 19.5 L}{278.5 \cancel{K}} \times \frac{338 \cancel{K}}{1.5 \cancel{a t m}} = 45.8 L$