The cyclic process for 1 mole of an ideal gas is shown in the V-T diagram. The work done in AB, BC and CA respectively is

Options

A. `0, RT_(2) 1n ((V_(1))/(V_(2))), R (T_(1) - T_(2))`
B. `R (T_(1) - T_(2)), 0, RT_(1) 1n ((V_(1))/(V_(2)))`
C. `0, RT_(2) 1n ((V_(2))/(V_(1))), R (T_(1) - T_(2))`
D. `0, RT_(2) 1n ((V_(2))/(V_(1))), R (T_(2) - T_(1))`

Answer

C

Explanation

c. Process `AB` is isochoric, therefore
`W_(AB) = P Delta V = 0`
Process `BC` isothermal, therefore
`W_(BC) = RT_(2) 1n ((V_(2))/(V_(1)))`
Process `CA` is isobaric, therefore
`W_(CA) = P Delta V = R Delta T = R (T_(2) - T_(1))`