If $\frac{7}{3} L$ $7/3 L$ of a gas at room temperature exerts a pressure of $24 k P a$ $24 kPa$ on its container, what pressure will the gas exert if the container's volume changes to $\frac{19}{4} L$ $19/4 L$ ?

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Answer 1 · 2016-12-11T07:51:50

The pressure is $= 11.8 k P a$ $=11.8 kPa$

We use Boyle's Law

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

Here,

$P_{1} = 24 k P a$ $P_1=24 kPa$

$V_{1} = \frac{7}{3} L$ $V_1=7/3L$

$V_{2} = \frac{19}{4} L$ $V_2=19/4L$

$P_{2} = \frac{P_{1} V_{1}}{V_{2}} = 24 \cdot \frac{7}{3} \cdot \frac{4}{19} = 11.8 k P a$ $P_2=(P_1V_1)/V_2=24*7/3*4/19=11.8kPa$

If 73L7/3 L of a gas at room temperature exerts a pressure of 24kPa24 kPa on its container, what pressure will the gas exert if the container's volume changes to 194L19/4 L?