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Question
Consider an isolated system and process as shown in the figure. (Process is that Seeta sitting on the freely movable piston jumps out). Assume no effect of reaction force due to jump on the system. Pressure equivalent of piston is 9 atm and the pressure equivalent when Seeta is on the piston is 10 atm. Atmospheric pressure is 1 atm. (given molar heat capacity at constant volume CV=3 Cal mol−1 K−1 )
Final temperature of the system in state - 2 is:
Consider an isolated system and process as shown in the figure. (Process is that Seeta sitting on the freely movable piston jumps out). Assume no effect of reaction force due to jump on the system. Pressure equivalent of piston is 9 atm and the pressure equivalent when Seeta is on the piston is 10 atm. Atmospheric pressure is 1 atm. (given molar heat capacity at constant volume CV=3 Cal mol−1 K−1 )
Final temperature of the system in state - 2 is:
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Solution
The correct option is D 47∘C
Process is adiabatic and irreversible as the expansion is sudden, against a constant pressure of 10 atm(p2)
By the first law of Thermodynamics
nCv(T2−T1)=w=−p2(v2−v1)
⇒nCv(T2−T1)=−p2[nRT2P2−nRT1P1]
3(T2−T1)=−10[RT210−RT120]
3(T2−T1)=−[2T2−T1]
⇒5T2=4T1
⇒T2=45T1 T2=45×400=320 K=47∘C
Process is adiabatic and irreversible as the expansion is sudden, against a constant pressure of 10 atm(p2)
By the first law of Thermodynamics
nCv(T2−T1)=w=−p2(v2−v1)
⇒nCv(T2−T1)=−p2[nRT2P2−nRT1P1]
3(T2−T1)=−10[RT210−RT120]
3(T2−T1)=−[2T2−T1]
⇒5T2=4T1
⇒T2=45T1 T2=45×400=320 K=47∘C
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