Question

70 calories of heat are required to raise the temperature of 2 moles of an ideal diatomic gas at constant from \[30^\circ C\] to \[35^\circ C\]. The amount of heat required (in calorie) to raise the temperature of the same gas through the same range (\[30^\circ C\] to \[35^\circ C\]) at constant volume is
A. 30
B. 50
C. 70
D. 90

Answer 1

Hint: calculate the specific heat at constant pressure using the formula for heat lost or gained by the gas at constant pressure. Determine the specific heat at constant volume using the relationship between specific heat at constant volume, specific heat at constant pressure and gas constant. Use the formula for heat lost or gained by the gas in isochoric process (constant volume process).

Formula used:
\[\Delta Q = n{C_P}\left( {{T_f} - {T_i}} \right)\]
Here, n is the number of moles of gas, \[{C_P}\] is the specific heat capacity at constant pressure, \[{T_f}\] is the final temperature and \[{T_i}\] is the initial temperature.
\[\Delta Q = n{C_V}\left( {{T_f} - {T_i}} \right)\]
Here, \[{C_V}\] is the specific heat capacity at constant volume.

Complete step by step answer:
We have given the heat required to raise the temperature of 2 moles of an ideal diatomic gas for the isobaric process, \[\Delta Q = 70\,{\text{cal}}\]. We have to determine the specific heat capacity at constant pressure.
We know the expression for heat lost or gained by the gas for isobaric process (constant pressure),
\[\Delta Q = n{C_P}\left( {{T_f} - {T_i}} \right)\]
Here, n is the number of moles of gas, \[{C_P}\] is the specific heat capacity at constant pressure, \[{T_f}\] is the final temperature and \[{T_i}\] is the initial temperature.
Substituting \[\Delta Q = 70\,{\text{cal}}\], \[n = 2\,{\text{mol}}\], \[{T_f} = 35^\circ C\] and \[{T_i} = 30^\circ C\] in the above equation, we get,
\[70 = \left( 2 \right){C_P}\left( {35 - 30} \right)\]
\[ \Rightarrow 70 = 10{C_P}\]
\[ \Rightarrow {C_P} = 7\,{\text{cal}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}\]
We know the relation between the specific heat at constant pressure and the specific heat at constant volume,
\[{C_P} - {C_V} = R\]

Here, \[{C_V}\] is the specific heat at constant volume and R is the gas constant.
Substituting \[{C_P} = 7\,{\text{cal}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}\] and \[R = 2\,{\text{cal}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}\,{{\text{K}}^{ - 1}}\] in the above equation, we get,
\[7 - {C_V} = 2\]
\[ \Rightarrow {C_V} = 5\,{\text{cal}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}\]
We know the expression for heat lost or gained by the gas for isochoric process (constant volume),
\[\Delta Q = n{C_V}\left( {{T_f} - {T_i}} \right)\]
Substituting \[{C_V} = 5\,{\text{cal mo}}{{\text{l}}^{ - 1}}\,{{\text{K}}^{ - 1}}\], \[n = 2\,{\text{mol}}\], \[{T_f} = 35^\circ C\] and \[{T_i} = 30^\circ C\] in the above equation, we get,
\[\Delta Q = \left( 2 \right)\left( 5 \right)\left( {35 - 30} \right)\]
\[ \Rightarrow \Delta Q = 50\,{\text{cal}}\]

So, the correct answer is “Option B”.

Note:
Students should note that the value of gas constant is \[8.31\,{\text{J}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}\,{{\text{K}}^{ - 1}}\] if the heat energy is given in joule and it is \[2\,{\text{cal}}\,\,{\text{mo}}{{\text{l}}^{ - 1}}\,{{\text{K}}^{ - 1}}\] if the heat energy is given in calories. Also, in the formula for heat lost or gained by the gas is the number of moles and not the number of molecules of the gas. This equation is valid for only gases.