Question #605aa

May 3, 2016

${V}_{2} = 9.06 m L$

Explanation:

Since the number of mole $n$ is constant and the factors changing are the volume, the pressure and temperature; we can modify the ideal gas law $P V = n R T$ to be:

$\frac{P V}{T} = k$ where $k = n R$ is constant.

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

$\implies {V}_{2} = \frac{{P}_{1} {V}_{1}}{{T}_{1}} \times \frac{{T}_{2}}{{P}_{2}} = \frac{0.975 \cancel{a t m} \times 8.88 m L}{261 \cancel{K}} \times \frac{273 \cancel{K}}{1 \cancel{a t m}} = 9.06 m L$

Therefore, the volume of the gas at STP would be ${V}_{2} = 9.06 m L$.