If $11/6 L$ of a gas at room temperature exerts a pressure of $9 kPa$ on its container, what pressure will the gas exert if the container's volume changes to $5/4 L$ ?

Question

1306 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-18T14:12:50

The pressure is $=13.2kPa$

We apply Boyle's Law

$P_1V_1=P_2V_2$

The initial pressure is $P_1=9kPa$

The initial volume is $V_1=11/6L$

The final volume is $V_2=5/4L$

The final pressure is

$P_2=V_1/V_2*P_1$

$=(11/6)/(5/4)*9$

$=13.2kPa$

If 11/6 L11/6 L of a gas at room temperature exerts a pressure of 9 kPa9 kPa on its container, what pressure will the gas exert if the container's volume changes to 5/4 L5/4 L?