# If 7/6 L of a gas at room temperature exerts a pressure of 6 kPa on its container, what pressure will the gas exert if the container's volume changes to 2/3 L?

Jan 14, 2018

As volume reduction process doesn't allow any change in mass or temperature of the gas,we can apply Boyle's law i.e $P V = c o n s \tan t$
$P 1 V 1 = P 2 V 2$ ($P 2$ is the pressure exerted by the gas on the container after volume reduction)
Or, $\left(\frac{7}{6}\right) 6$ = $P 2 \cdot \left(\frac{2}{3}\right)$
Or, $P 2$ = $\frac{21}{2} k P a$ or $10.5 k P a$