Question

A \[0.145{\text{ d}}{{\text{m}}^{\text{3}}}\] of hydrogen gas is collected over water at \[23^\circ {\text{C}}\] and total pressure of \[{\text{99085 N}}{{\text{m}}^{{\text{ - 2}}}}\]. If the vapour pressure of water at \[23^\circ {\text{C}}\] is \[{\text{2973 N}}{{\text{m}}^{{\text{ - 2}}}}\]. What is the volume of dry gas under NTP conditions?
A. 0.136
B. 0.145
C. 0.169
D. 0.192

Answer 1

Hint: Using the given total pressure of gas and vapour pressure of water calculate the pressure of dry hydrogen gas at \[23^\circ {\text{C}}\]. Using the combined gas law, calculate the dry hydrogen gas under NTP conditions.

Formula Used:
\[{\text{Total pressure = Water vapor pressure + Pressure of dry hydrogen gas}}\]
Combined gas law: \[\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}\]

Complete answer:
We have given total pressure of gas and vapour pressure of water at \[23^\circ {\text{C}}\].
Total pressure = \[{\text{99085 N}}{{\text{m}}^{{\text{ - 2}}}}\]
Vapour pressure of water = \[{\text{2973 N}}{{\text{m}}^{{\text{ - 2}}}}\]
Using the total pressure of gas and vapour pressure of water we have to calculate the pressure of dry hydrogen gas at \[23^\circ {\text{C}}\].
\[{\text{Total pressure = Water vapor pressure + Pressure of dry hydrogen gas}}\]
Substitute \[{\text{99085 N}}{{\text{m}}^{{\text{ - 2}}}}\] for total pressure, \[{\text{2973 N}}{{\text{m}}^{{\text{ - 2}}}}\] for water vapour pressure and calculate the pressure of dry hydrogen gas.
\[\Rightarrow {\text{99085 N}}{{\text{m}}^{{\text{ - 2}}}} = {\text{2973 N}}{{\text{m}}^{{\text{ - 2}}}}{\text{ + Pressure of dry hydrogen gas}}\]
\[\Rightarrow {\text{Pressure of dry hydrogen gas = 99085 N}}{{\text{m}}^{{\text{ - 2}}}} - {\text{2973 N}}{{\text{m}}^{{\text{ - 2}}}} = 96112{\text{N}}{{\text{m}}^{{\text{ - 2}}}}\]
Now using the combined gas law we can calculate the volume of dry gas under NTP conditions.
\[\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}\]
\[{P_1}\] = \[96112{\text{N}}{{\text{m}}^{{\text{ - 2}}}}\]
\[{V_1}\] = \[0.145{\text{ d}}{{\text{m}}^{\text{3}}}\]
\[{T_1} = 23^\circ {\text{C}} = {\text{ (23 + 273)K = 296 K}}\]
At NTP condition pressure is 1 atm and temperature is \[20^\circ {\text{C}}\].
\[{P_2} = 1{\text{ atm = 101325 N}}{{\text{m}}^{{\text{ - 2}}}}\]
\[{T_2} = {\text{ 20}}^\circ {\text{C}} = {\text{ 293 K}}\]
\[\Rightarrow \dfrac{{96112{\text{N}}{{\text{m}}^{{\text{ - 2}}}} \times 0.145{\text{ d}}{{\text{m}}^{\text{3}}}}}{{{\text{296 K}}}} = \dfrac{{{\text{101325 N}}{{\text{m}}^{{\text{ - 2}}}}{{ \times }}{{\text{V}}_{\text{2}}}}}{{{\text{293K}}}}\]
\[\Rightarrow {V_2} = 0.136{\text{ d}}{{\text{m}}^{\text{3}}}\]
Thus, the volume of dry hydrogen gas under NTP conditions is \[0.136{\text{ d}}{{\text{m}}^{\text{3}}}\].

Hence, option (A) \[0.136{\text{ d}}{{\text{m}}^{\text{3}}}\] is the correct answer.

Note:
Water vapour pressure varies with temperature. When gas collected over water the total pressure is the sum of the vapour pressure of water and pressure of the gas.