# A 12.0 L sample of gas is at STP. What would be its new volume if its pressure was decreased to 575 mmHg and its temperature was doubled?

May 5, 2016

${V}_{2} = 31.7 L$

#### Explanation:

Let us first list the given data:

Position 1:
STP means that $P = 1 a t m = 760 m m H g$ and $T = 273 K$.
The volume occupied by the gas is $V = 12.0 L$.

Position 2:
The new conditions are:
$P = 575 m m H g$ and $T = 2 \times 273 = 546 K$.
V=?

To find the volume of the gas at position 2, we can rearrange the ideal gas law: $P V = n R T$ to be:

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

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

$\implies {V}_{2} = \frac{{P}_{1} {V}_{1}}{{T}_{1}} \times \frac{{T}_{2}}{{P}_{2}} = \frac{760 \cancel{m m H g} \times 12.0 L}{273 \cancel{K}} \times \frac{546 \cancel{K}}{575 \cancel{m m H g}} = 31.7 L$