Solveeit Logo
Login

Question

Question: Work done in reversible adiabatic process by an ideal gas is given by: A.2.303 RT log \(\dfrac{{{V...

Work done in reversible adiabatic process by an ideal gas is given by:
A.2.303 RT log V1V2\dfrac{{{V_1}}}{{{V_2}}}
B.None
C.2.303 RT log V2V1\dfrac{{{V_2}}}{{{V_1}}}​​
D.nR(γ−1)(T2−T1)\dfrac{{nR}}{{(\gamma - 1)}}(T2 - T1)

Explanation

Solution

An adiabatic process is one in which no heat enters or leaves the system, and hence, for a reversible adiabatic process the first law takes the form dU=−PdVdU = - PdV from the equation CV=(dU/dT)V{C_V} = {\left( {dU/dT} \right)_V}. But the internal energy of an ideal gas depends only on the temperature and is independent of the volume because there are no intermolecular forces, and hence, for an ideal gas, CV=dU/dT{C_V} = dU/dT, and so we have the equation as dU=CVdTdU = {C_V}dT. Thus, for a reversible adiabatic process and an ideal gas, CVdT=−PdV{C_V}dT = - PdV and the minus sign here shows that as V increases, T decreases.

Complete step by step answer:
The adiabatic process can be derived from the first law of thermodynamics relating to the change in internal energy dU to the work W done by the system and the heat dQ added to it.
From first law of thermodynamics, we know that
ΔU=q+W\Delta U = q + W
Where,
= Change in internal energy
q = heat absorbed or released
w = work done on or by the system
As we know that, for an adiabatic process,
q = 0
∴ΔU=W\therefore \Delta U = W ……………equation 1
As we know that,
ΔU=nCvΔT\Delta U = nCv\Delta T
whereas,
ΔT\Delta T = Change in temperature =T2−T1 = {T_2}-{T_1}
Cv ​= heat capacity at constant volume =R(γ−1) = \dfrac{R}{{(\gamma - 1)}}
n = no. of moles
∴ΔU=nR(γ−1)(T2−T1)\therefore \Delta U = n\dfrac{R}{{(\gamma - 1)}}(T2 - T1)………………equation 2
From equation (1) and (2), we have
W=nR(γ−1)(T2−T1)W = \dfrac{{nR}}{{(\gamma - 1)}}(T2 - T1)
Hence, work done in reversible adiabatic process by an ideal gas is given bynR(γ−1)(T2−T1)\dfrac{{nR}}{{(\gamma - 1)}}(T2 - T1)

Therefore, the correct answer is option (D).
Note:
We should note also that, since R=  CP  −  CVR = \;{C_P}\; - \;{C_V} and CP/CV=γ{C_P}/{C_V} = \gamma this can also be written as W=Cv(T1−T2)W = {C_v}\left( {{T_1}-{T_2}} \right). This is also equal to the heat that would be lost if the gas were to cool from T1{T_1} to T2{T_2} at constant volume.