\(200 cm ^{3}\) of hydrogen were collected over water at \(30^{\circ} C\) and \(740 mm\) of \(Hg\). Calculate the volume of the gas at s.t.p. If the vapour pressure of water at the temperature of the experiment is \(14 mm\) of \(Hg .
Explanation
When a gas is collected over water, the measured pressure (total pressure) is the sum of the gas pressure and that of water vapour. That is
Measured pressure
\(=\left(\begin{array}{c}\text { pressure } \\ \text { of dry gas }\end{array}\right)+\left(\begin{array}{c}\text { pressure due to } \\ \text { water vapour }\end{array}\right)\)
Measured pressure \(\left(\right.\) at \(\left.30^{\circ} C \right)\)
\(=740 mmHg\)
Vapour pressured \(\left(\right.\) at \(\left.30^{\circ} C \right)\)
\(=14 mmHg\)
Volume of hydrogen, \(V_{1}=200 cm ^{3}\)
Pressure of dry gas
= measured pressure - pressure due to water vapour
\(=740 mmHg -14 mmHg\)
\(=726 mmHg\)
\(T_{1}=30^{\circ} C =(273.15+30) K\)
\(=303.15 K\)t STP (standard temperature \& pressure)
\(P_{0}=760 mmHg , V_{0}=? T_{0}=273.15 K\)
Using the gencral gas cquation,
\begin{aligned}
&\frac{P_{0} V_{0}}{T_{0}}=\frac{P_{1} V_{1}}{T_{1}} ; V_{0}=\frac{P_{1} V_{1} T_{0}}{P_{0} T_{1}} \\
&=\frac{726 mmHg \times 200 cm ^{3} \times 273.15 K }{760 mmHg \times 303.15 K } \\
&=172.14 cm ^{3}
\end{aligned}