# Question #e9969

Dec 4, 2014

The final pressure will be ${P}_{2} = 0.39 a t m$.

Using the ideal gas law $P V = n R T$ and knowing that your initial state is defined by a volume ${V}_{1} = 225 m L$, a pressure ${P}_{1} = 1 a t m$, and a temperature of $T = \left(0 + 273.15\right) = 273.15 K$, you can determine that

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

keeping in mind that the number of moles, $n$, remains constant. Your final values for temperature and volume are

${V}_{2} = 625 m L$ and ${T}_{2} = \left(25 + 273.15\right) = 298.15 K$, respectively.

Therefore,

${P}_{2} = \frac{{V}_{1}}{{V}_{2}} \cdot \frac{{T}_{2}}{{T}_{1}} \cdot {P}_{1}$ , which yields ${P}_{2} = 0.39 a t m$.

The answer makes sense, since an increasing volume would be correlated with a drop in pressure, given that the number of moles are constant.