# If 9/7 L of a gas at room temperature exerts a pressure of 12 kPa on its container, what pressure will the gas exert if the container's volume changes to 3/4 L?

Mar 6, 2016

Gas will exert a pressure of $20.47 k P a$

#### Explanation:

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

As no change in the temperature has been mentioned it is assumed to be constant at room temperature and then ${P}_{1} \times {V}_{1} = {P}_{2} \times {V}_{2}$

In the given case, $\frac{9}{7}$ $L$ of gas inside of a container exerts a pressure of $12 k P a$ of pressure and is at a temperature of ${30}^{o} K$. When the volume of the gas changes to $\frac{3}{4}$ $L$, the pressure becomes ${P}_{2}$, we should have

$12 \times \frac{9}{7} = {P}_{2} \times \frac{3}{4}$ or ${P}_{2} = \frac{12 \times 9 \times 4}{3 \times 7} = \frac{144}{7} = 20.47$

i.e. ${P}_{2} = 20.47 k P a$