If $8 L$ $8 L$ of a gas at room temperature exerts a pressure of $15 k P a$ $15 kPa$ on its container, what pressure will the gas exert if the container's volume changes to $15 L$ $15 L$ ?

Question

1382 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-26T12:08:34

The pressure is $= 8 k P a$ $=8kPa$

We apply Boyle's Law

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$

The initial volume is $V_{1} = 8 L$ $V_1=8L$

The initial pressure is $P_{1} = 15 k P a$ $P_1=15kPa$

The final volume is $V_{2} = 15 L$ $V_2=15L$

The final pressure

$P_{2} = \frac{V_{1}}{V_{2}} \cdot P_{1}$ $P_2=V_1/V_2*P_1$

$= \frac{8}{15} \cdot 15$ $=8/15*15$

$= 8 k P a$ $=8kPa$

If 8 L8L8 L of a gas at room temperature exerts a pressure of 15 kPa15kPa15 kPa on its container, what pressure will the gas exert if the container's volume changes to 15 L15L15 L?