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Two moles of helium gas are taken along the path `ABCD` (as shown in Fig.) The work done by the gas is A. `2000R ((1)/(2) + 1n (4)/(3))`B. `500 R (R + 1n 4)`C. `500 R (2 + 1n (16)/(9))`D. `1000 R (1 + 1n (16)/(9))` |
Answer» Correct Answer - A a. `A rarr B` is isobaric process, `V prop T` so, `Delta W_(AB) = nR Delta T = 2 xx R xx (750 - 250) = 1000 R` `B rarr C` is an isochoric process `:. Delta W_(BC) = 0` and `C rarr D` is an isothermal process `Delta W_(CD) = nRT 1n ((V_(f))/(V_(i)))` ` = 2 xx R xx 1000 1n ((20)/(15)) = 2000 1n ((4)/(3))` Total work done, `Delta W = Delta W_(AB) + Delta W_(BC) + Delta W_(CD)` |
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