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Question: One mole of an ideal gas \[\left( {{C_{v,m}} = \dfrac{5}{2}R} \right)\] at \[300{\text{ }}k\]and\[5{...

One mole of an ideal gas (Cv,m=52R)\left( {{C_{v,m}} = \dfrac{5}{2}R} \right) at 300 k300{\text{ }}kand5 atm5{\text{ }}atmis expanded adiabatically to a final Pressure of 2 atm2{\text{ }}atmagainst a constant pressure of2 atm2{\text{ }}atm. Final temperature of the gas is:
(A) 270K270K
(B) 273K273K
(C) 248.5K248.5K
(D) 200K200K

Explanation

Solution

students remember that in adiabatic process work done is equal to change in internal energy. So use the formula for work done in terms of temperature to find the final temperature of the system,work done is expressed in terms of temperature. We use the first law of thermodynamics. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.

Complete step by step answer:
The parameters given in the question are:
Number of mole(n) = 1 mole\left( n \right){\text{ }} = {\text{ }}1{\text{ }}mole
Initial pressure(P1) = 5 atm\left( {{P_1}} \right){\text{ }} = {\text{ }}5{\text{ }}atm
Final pressure(P2) = 2 atm\left( {{P_2}} \right){\text{ }} = {\text{ }}2{\text{ }}atm
External pressure(Pext) = 2 atm\left( {{P_{ext}}} \right){\text{ }} = {\text{ }}2{\text{ }}atm
Initial temperature(T1) = 300 k\left( {{T_1}} \right){\text{ }} = {\text{ }}300{\text{ }}k
Cv,m=52R{C_{v,m}} = \dfrac{5}{2}R
We have to find final temp (T2)
In an adiabatic process
ΔQ = 0\Delta Q{\text{ }} = {\text{ }}0
By first law of thermodynamics
ΔQ=ΔU+ΔW ΔU+ΔW=0 nCvΔT+ΔW=0 ΔW=−nCvΔT   \Delta Q = \Delta U + \Delta W \\\ \Delta U + \Delta W = 0 \\\ n{C_v}\Delta T + \Delta W = 0 \\\ \Delta W = - n{C_v}\Delta T \\\ \\\
Also,
ΔW=−PExtΔV PExtΔV=−nCvΔT PExt(V2−V1)=−nCv(T2−T1) PExt(nRT2P2−nRT1P1)=−nCv(T2−T1) PExtR(T2P2−T1P1)=−Cv(T2−T1) 2R(T22−3005)=−52R(T2−300) 2(T22−60)=−52(T2−300) T2−120=−52T2+750 T2+52T2=120+750 72T2=870 T2=248.5K  \Delta W = - {P_{Ext}}\Delta V \\\ {P_{Ext}}\Delta V = - n{C_v}\Delta T \\\ {P_{Ext}}({V_2} - {V_1}) = - n{C_v}({T_2} - {T_1}) \\\ {P_{Ext}}\left( {\dfrac{{nR{T_2}}}{{{P_2}}} - \dfrac{{nR{T_1}}}{{{P_1}}}} \right) = - n{C_v}({T_2} - {T_1}) \\\ {P_{Ext}}R\left( {\dfrac{{{T_2}}}{{{P_2}}} - \dfrac{{{T_1}}}{{{P_1}}}} \right) = - {C_v}({T_2} - {T_1}) \\\ 2R\left( {\dfrac{{{T_2}}}{2} - \dfrac{{300}}{5}} \right) = - \dfrac{5}{2}R({T_2} - 300) \\\ 2\left( {\dfrac{{{T_2}}}{2} - 60} \right) = - \dfrac{5}{2}({T_2} - 300) \\\ {T_2} - 120 = - \dfrac{5}{2}{T_2} + 750 \\\ {T_2} + \dfrac{5}{2}{T_2} = 120 + 750 \\\ \dfrac{7}{2}{T_2} = 870 \\\ {T_{_2}} = 248.5K \\\
Hence, the correct option is (C).

Note: Some important types of thermodynamic process are :
1)Isochoric process : In an isochoric process the volume of the system remains constant. Hence the work done by the system is zero.
2)Isobaric process : In an Isobaric process the pressure of the system remains constant.
3)Isothermal process : In an Isothermal process the temperature of the system remains constant throughout the process. Isothermal process is carried out very slowly so that there is a heat exchange between the system and the surrounding so that temperature can remain constant .
4)Adiabatic process : In an Adiabatic process there is no heat exchange between the system and the surroundings.
In adiabatic processes there is no heat exchange between the system and surroundings . Hence by the first law of thermodynamics the work is done in terms of internal energy.