# A container with a volume of 6 L contains a gas with a temperature of 620^o K. If the temperature of the gas changes to 370 ^o K without any change in pressure, what must the container's new volume be?

Dec 21, 2016

Container's new volume will be $3.58 L$.

#### Explanation:

When a fixed quantum (mass) of gas in a container under goes changes in its pressure, volume and temperature, $\frac{P V}{T}$ always remains constant, where $p$ is pressure , $V$ is volume and $T$ is its temperature in degree Kelvins.

Here there is no change n pressure, hence $\frac{V}{T}$ is constant i.e. ${V}_{i} / {T}_{i} = {V}_{f} / {T}_{f}$,

where ${V}_{i}$ and ${T}_{i}$ are initial volume and temperature and ${V}_{f}$ and ${T}_{f}$ are final volume and temperature.

As ${V}_{i} = 6 L$ and ${T}_{i} = {620}^{\circ}$ $K$ and ${T}_{f} = {370}^{\circ}$$\mathmr{and}$V_f# is given by

$\frac{6}{620} = {V}_{f} / 370$ and ${V}_{f} = \frac{6 \times 370}{620} = 3.58 L$