A carnot engine has a cycle as given in figure. If W1 and W2 represent work done by one mole of monoatomic and diatomic gas respectively in the complete cyclic process, then calculate W1W2?

n=1−T2T1=1−14=34For mono atomic gas34=W1q12T2V2r−1=T3V3r−1V2V3r−1=T3T1=14V264V053−1=14V264V023=14V264V0=1432=18V2=8V0q12=−W12=nRT2 ln8W1=3nRT2ln⁡2×34 For diatomic gas V2V3r−1=T3T1=14V2V32/5=14V2=145/2V3V2=132×64V0V2=2V0q12=−W12=nRT2ln⁡2W2=34q12=34×nRT2ln⁡2W1W2=3