Question

The temperature of 3 kg of nitrogen is raised from \[10^\circ \;C\] to \[100^\circ \;C\] . Compute the difference in work done if heating were at constant pressure and at constant volume.
( For nitrogen \[{C_p} = 1400Jk{g^{ - 1}}{K^{ - 1}}\] & \[{C_V} = 740Jk{g^{ - 1}}{K^{ - 1}}\])
A. 81000 J
B. 64000 J
C. 49000 J
D. 36000 J

Answer 1

Hint:-First find the number of moles in 3kg of nitrogen , then use the formula work done to raise the temperature of n number of moles by \[\Delta T\]at constant pressure (\[W = n{C_p}\Delta T\]) and also for work done to raise the temperature of n number of moles by \[\Delta T\]at constant volume (\[W = n{C_V}\Delta T\]).
After finding all the differences by subtracting , this is the required answer.

Complete step-by-step solution:
Work done to raise the temperature from \[10^\circ \;C\] to \[100^\circ \;C\]at constant pressure \[W = n{C_p}\Delta T\]
Work done to raise the temperature from \[10^\circ \;C\] to \[100^\circ \;C\]at constant volume \[W = n{C_V}\Delta T\]
Since in both the cases the temperature is increasing by the same degree so change in internal energy for both the cases is the same.
Difference in work done if heating were at constant pressure and at constant volume is given by \[\Delta W\]
\[ \Rightarrow \Delta W = n{C_p}\Delta T - n{C_V}\Delta T\]
By simplifying the relation we get,
\[ \Rightarrow \Delta W = n({C_p} - {C_v})\Delta T\]
We know that \[({C_p} - {C_v}) = R\]
\[ \Rightarrow ({C_p} - {C_v}) = R\]
\[ \Rightarrow R = 8.314Jk{g^{ - 1}}{K^{ - 1}}\]
\[ \Rightarrow number\;of\;moles = \dfrac{{weight}}{{molar\;mass}}\]
Finding the numbers of moles of nitrogen in 3kg.
\[ \Rightarrow n = \dfrac{{3kg}}{{28gm}} = \dfrac{{3000gm}}{{28gm}} = 107.142moles\]
\[ \Rightarrow \Delta W = n({C_p} - {C_v})\Delta T\] \[\because ({C_p} - {C_v}) = R\] & \[R = 8.314Jk{g^{ - 1}}{K^{ - 1}}\]
Putting all the values we get,
\[ \Rightarrow \Delta W = 107.142(8.314).(100 - 10)\]
By solving this we get with
\[ \Rightarrow \Delta W \approx 81000J\]
Since , the difference in work done if heating were at constant pressure and at constant volume is \[81000J\].

Hence option (A) is the correct answer.

Note:- Work done to increase the temperature of gases at constant pressure is always greater than the work done to increase the temperature by the same amount of gases at constant volume.
However, in both the cases the temperature is increasing by the same degree so change in internal energy for both the cases is the same.