Question

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`?

Answer 1

Answer is 13.067 kPa

Explanation:

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