If 7/6 L of a gas at room temperature exerts a pressure of 2 kPa on its container, what pressure will the gas exert if the container's volume changes to 5/4 L?

Feb 24, 2018

The pressure will be $= 1.87 k P a$

Explanation:

Apply Boyle's Law

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, at constant temperature

The initial pressure is ${P}_{1} = 2 k P a$

The initial volume is ${V}_{1} = \frac{7}{6} L$

The final volume is ${V}_{2} = \frac{5}{4} L$

The final pressure is

${P}_{2} = {V}_{1} / {V}_{2} \cdot {P}_{1} = \frac{\frac{7}{6}}{\frac{5}{4}} \cdot 2 = 1.87 k P a$