At $25^{o} C$ and $760 m m$ of $H g$ pressure, a gas occupies $600 m L$ volume. What will be its pressure at a height where temperature is $10^{o} C$ and volume of the gas is $640 m L$ ?

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Solution

Here, $P_{1} = 760 m m H g; V_{1} = 600 m l; T_{1} = 298 K$
And, $V_{2} = 640 m l; T_{2} = 283 K$
According to combined gas law equation, $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$P_{2} = \frac{P_{1} V_{1} T_{2}}{V_{2} T_{1}}$
$= \frac{760 \times 600 \times 283}{640 \times 298}$
$= 676.6 m m H g$

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At 25oC and 760 mm of Hg pressure, a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10oC and volume of the gas is 640 mL?

Here, P1=760mmHg;V1=600ml;T1=298KAnd, V2=640ml;T2=283KAccording to combined gas law equation, P1V1T1=P2V2T2P2=P1V1T2V2T1=760×600×283640×298=676.6mmHg

At $25^{o} C$ and $760 m m$ of $H g$ pressure, a gas occupies $600 m L$ volume. What will be its pressure at a height where temperature is $10^{o} C$ and volume of the gas is $640 m L$ ?