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One mole of an ideal monatomic gas starts at temperature T0, pressure P0, and volume V0 and undergoes four reversible steps: 1. Isobaric expansion until temperature reaches (1+beta)*T0. 2. Isothermal expansion until pressure drops to P0/alpha. 3. Isobaric contraction until temperature returns to T0. 4. Isothermal contraction until pressure returns to P0. Which of the following statements are correct?

1. Total heat supplied to the gas equals (5/2)*beta*P0*V0
2. Total heat rejected by the gas equals beta*P0*V0
3. Total work done by the gas equals P0*V0*beta*ln(alpha)
4. Efficiency of the cycle equals beta*ln(alpha) / ((1+beta)*ln(alpha) + 5*beta/2)

Correct answer: Total heat supplied to the gas equals (5/2)*beta*P0*V0

Solution

Step 1 (isobaric, P0): Q1 = (5R/2)*beta*T0 = (5/2)*beta*P0*V0. This is the total heat supplied in step 1. Step 2 (isothermal, T=(1+beta)T0): Q2 = (1+beta)*P0*V0*ln(alpha) (supplied). Step 3 (isobaric, P0/alpha): Q3 = (5/2)*beta*P0*V0 (rejected). Step 4 (isothermal, T=T0): Q4 = P0*V0*ln(alpha) (rejected). Total heat supplied = (5/2)*beta*P0*V0 + (1+beta)*P0*V0*ln(alpha). Total rejected = (5/2)*beta*P0*V0 + P0*V0*ln(alpha). Net work = beta*P0*V0*ln(alpha). Options A, C, D are correct; B is incorrect.

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