At $25^{o} C$ and 760 mm of Hg pressure, a gas occupies 600 mL volume. What will be its pressure at a height where temperatures is $10^{o} C$ and volume of the gas is 640 mL?

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Solution

Given, $P_{1} = 760 m m H g$ , $V_{1} = 600 m L$

$T_{1} = 273 + 25 = 298 K$

$P_{2} = ?$
$V_{2} = 640 m L$ and $T_{2} = 273 + 10 = 283 K$

According to combined gas law equation,

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

or $P_{2} = \frac{P_{1} V_{1} T_{2}}{V_{2} T_{1}}$

$= \frac{(760 m m H g) \times (600 m L) \times (283 K)}{(640 m L) (298 K)}$

= $676.6 m m H g$

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