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Question: A \[0.145{\text{ d}}{{\text{m}}^{\text{3}}}\] of hydrogen gas is collected over water at \[23^\circ ...

A 0.145 dm30.145{\text{ d}}{{\text{m}}^{\text{3}}} of hydrogen gas is collected over water at 23∘C23^\circ {\text{C}} and total pressure of 99085 Nm - 2{\text{99085 N}}{{\text{m}}^{{\text{ - 2}}}}. If the vapour pressure of water at 23∘C23^\circ {\text{C}} is 2973 Nm - 2{\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

Explanation

Solution

Using the given total pressure of gas and vapour pressure of water calculate the pressure of dry hydrogen gas at 23∘C23^\circ {\text{C}}. Using the combined gas law, calculate the dry hydrogen gas under NTP conditions.

Formula Used:
Total pressure = Water vapor pressure + Pressure of dry hydrogen gas{\text{Total pressure = Water vapor pressure + Pressure of dry hydrogen gas}}
Combined gas law: P1V1T1=P2V2T2\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∘C23^\circ {\text{C}}.
Total pressure = 99085 Nm - 2{\text{99085 N}}{{\text{m}}^{{\text{ - 2}}}}
Vapour pressure of water = 2973 Nm - 2{\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∘C23^\circ {\text{C}}.
Total pressure = Water vapor pressure + Pressure of dry hydrogen gas{\text{Total pressure = Water vapor pressure + Pressure of dry hydrogen gas}}
Substitute 99085 Nm - 2{\text{99085 N}}{{\text{m}}^{{\text{ - 2}}}} for total pressure, 2973 Nm - 2{\text{2973 N}}{{\text{m}}^{{\text{ - 2}}}} for water vapour pressure and calculate the pressure of dry hydrogen gas.
⇒99085 Nm - 2=2973 Nm - 2 + 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}}
⇒Pressure of dry hydrogen gas = 99085 Nm - 2−2973 Nm - 2=96112Nm - 2\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.
P1V1T1=P2V2T2\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}
P1{P_1} = 96112Nm - 296112{\text{N}}{{\text{m}}^{{\text{ - 2}}}}
V1{V_1} = 0.145 dm30.145{\text{ d}}{{\text{m}}^{\text{3}}}
T1=23∘C= (23 + 273)K = 296 K{T_1} = 23^\circ {\text{C}} = {\text{ (23 + 273)K = 296 K}}
At NTP condition pressure is 1 atm and temperature is 20∘C20^\circ {\text{C}}.
P2=1 atm = 101325 Nm - 2{P_2} = 1{\text{ atm = 101325 N}}{{\text{m}}^{{\text{ - 2}}}}
T2= 20∘C= 293 K{T_2} = {\text{ 20}}^\circ {\text{C}} = {\text{ 293 K}}
⇒96112Nm - 2×0.145 dm3296 K=101325 Nm - 2×V2293K\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}}}}
⇒V2=0.136 dm3\Rightarrow {V_2} = 0.136{\text{ d}}{{\text{m}}^{\text{3}}}
Thus, the volume of dry hydrogen gas under NTP conditions is 0.136 dm30.136{\text{ d}}{{\text{m}}^{\text{3}}}.

**Hence, option (A) 0.136 dm30.136{\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.