# If 7/3 L of a gas at room temperature exerts a pressure of 21 kPa on its container, what pressure will the gas exert if the container's volume changes to 15/4 L?

Temperature remaining constant, product of pressure and volume is always constant. It is observed that volume is $\frac{7}{3}$ L of a gas at room temperature exerts a pressure of 21kPa, hence product is 49. So, if volume changes to $\frac{15}{4}$ L, pressure would be $\frac{49}{\frac{15}{4}}$ i.e. 13.067 kPa.