If $\frac{5}{7} L$ $5/7 L$ of a gas at room temperature exerts a pressure of $9 k P a$ $9 kPa$ on its container, what pressure will the gas exert if the container's volume changes to $\frac{8}{9} L$ $8/9 L$ ?

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Answer 1 · 2017-03-07T16:08:34

The pressure is $= 7.23 k P a$ $=7.23kPa$

We apply Boyle's Law

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

$P_{1} = 9 k P a$ $P_1=9kPa$

$V_{1} = \frac{5}{7} L$ $V_1=5/7L$

$V_{2} = \frac{8}{9} L$ $V_2=8/9L$

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

$= \frac{\frac{5}{7}}{\frac{8}{9}} \cdot 9$ $=(5/7)/(8/9)*9$

$= \frac{5}{7} \cdot \frac{9}{8} \cdot 9$ $=5/7*9/8*9$

$= \frac{405}{56}$ $=405/56$

$= 7.23 k P a$ $=7.23kPa$

If 5/7 L57L5/7 L of a gas at room temperature exerts a pressure of 9 kPa9kPa9 kPa on its container, what pressure will the gas exert if the container's volume changes to 8/9 L89L8/9 L?