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Two moles of an ideal monatomic gas is t...

Two moles of an ideal monatomic gas is taken through a cycle ABCA as shown in the P-T diagram. During the process AB , pressure and temperature of the gas very such that `PT=Constant`. It `T_1=300K`, calculate

(a) the work done on the gas in the process AB and
(b) the heat absorbed or released by the gas in each of hte processes. Give answer in terms of the gas constant R.

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Verified by Experts

The correct Answer is:
A, B, C

For `PV^x=Constt.,` Molar heat capacity
`C=1/R/(gamma-1)+R/(1-x)=R/(5/3-1)+R/(1-1/2)`
Here `P^2V=constant of PV^(1//2)=constant`
`:. x=1/2`
`rArrC=3.5R`
`Q_(AtoB)=nCDeltaT=2(3.5R)(300-600)=-2100R`
Process B -C : Process is isobaric therefore
`Q_(BtoC)=nCDeltaT=(2)(5/2R)(T_C-T_B)`
`=2(5/2R)(2T_1-T_1)=(5R)(600-300)=1500R`
Heat is absorbed
Process C-A: Process is isothermal
`DeltaT=0 and Q_(CtoA)=W_(CtoA)=nRT_CIn((P_C)/(P_A))`
`=nR(2T_1)In((2P_1)/(P_1))=(2)(R)(600)In(2)=1200Rxx0.6932`
`Q_(CtoA)=831.6R(absorbed)`
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