# The gas inside of a container exerts 5 Pa of pressure and is at a temperature of 190 ^o K. If the pressure in the container changes to 32 Pa with no change in the container's volume, what is the new temperature of the gas?

Mar 6, 2016

New temperature of the gas is ${1216}^{o} K$

#### Explanation:

Pressure $P$, volume $V$ and temperature $T$ (in Kelvin) have the relation $P \frac{V}{T} = k$ where $k$ is a constant.

As there is no change in the container's volume, it means that ${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

In the given case, gas inside of a container exerts $5 P a$ of pressure and is at a temperature of ${190}^{o} K$. When pressure of the gas changes to $32 P a$, if the temperature becomes ${T}_{2}$, we should have

$\frac{5}{190} = \frac{32}{T} _ 2$ or ${T}_{2} = \left(\frac{190 \times 32}{5}\right)$

i.e. ${P}_{2} = \frac{6080}{5} = {1216}^{o} K$.