The gas inside of a container exerts 15 Pa of pressure and is at a temperature of 450 ^o K. If the pressure in the container changes to 25 Pa with no change in the container's volume, what is the new temperature of the gas?

Apr 25, 2018

${p}_{1} = 15 P a$
T_1=450°C
${p}_{2} = 25 P a$
V="const.

Explanation:

$\frac{{p}_{1} {V}_{1}}{T} _ 1 = \frac{{p}_{2} {V}_{2}}{{T}_{2}}$
Considering $V = \text{const.} \implies {V}_{1} = {V}_{2}$, it become
$\frac{{p}_{1}}{T} _ 1 = \frac{{p}_{2}}{{T}_{2}}$
${T}_{2} = {p}_{2} / {p}_{1} {T}_{2} = \frac{15 P a}{25 P a} \cdot \left(450 + 273\right) K = 434 K$