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Question: The temperature of 3 kg of nitrogen is raised from \[10^\circ \;C\] to \[100^\circ \;C\] . Compute t...

The temperature of 3 kg of nitrogen is raised from 10  C10^\circ \;C to 100  C100^\circ \;C . Compute the difference in work done if heating were at constant pressure and at constant volume.
( For nitrogen Cp=1400Jkg1K1{C_p} = 1400Jk{g^{ - 1}}{K^{ - 1}} & CV=740Jkg1K1{C_V} = 740Jk{g^{ - 1}}{K^{ - 1}})
A. 81000 J
B. 64000 J
C. 49000 J
D. 36000 J

Explanation

Solution

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 ΔT\Delta Tat constant pressure (W=nCpΔTW = n{C_p}\Delta T) and also for work done to raise the temperature of n number of moles by ΔT\Delta Tat constant volume (W=nCVΔTW = 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  C10^\circ \;C to 100  C100^\circ \;Cat constant pressure W=nCpΔTW = n{C_p}\Delta T
Work done to raise the temperature from 10  C10^\circ \;C to 100  C100^\circ \;Cat constant volume W=nCVΔTW = 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 ΔW\Delta W
ΔW=nCpΔTnCVΔT\Rightarrow \Delta W = n{C_p}\Delta T - n{C_V}\Delta T
By simplifying the relation we get,
ΔW=n(CpCv)ΔT\Rightarrow \Delta W = n({C_p} - {C_v})\Delta T
We know that (CpCv)=R({C_p} - {C_v}) = R
(CpCv)=R\Rightarrow ({C_p} - {C_v}) = R
R=8.314Jkg1K1\Rightarrow R = 8.314Jk{g^{ - 1}}{K^{ - 1}}
number  of  moles=weightmolar  mass\Rightarrow number\;of\;moles = \dfrac{{weight}}{{molar\;mass}}
Finding the numbers of moles of nitrogen in 3kg.
n=3kg28gm=3000gm28gm=107.142moles\Rightarrow n = \dfrac{{3kg}}{{28gm}} = \dfrac{{3000gm}}{{28gm}} = 107.142moles
ΔW=n(CpCv)ΔT\Rightarrow \Delta W = n({C_p} - {C_v})\Delta T (CpCv)=R\because ({C_p} - {C_v}) = R & R=8.314Jkg1K1R = 8.314Jk{g^{ - 1}}{K^{ - 1}}
Putting all the values we get,
ΔW=107.142(8.314).(10010)\Rightarrow \Delta W = 107.142(8.314).(100 - 10)
By solving this we get with
ΔW81000J\Rightarrow \Delta W \approx 81000J
Since , the difference in work done if heating were at constant pressure and at constant volume is 81000J81000J.

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.