One mole of a monoatomic ideal gas is taken through the cycle ABCDA as shown in the figure.
`T_A=1000K` and `2p_A=3p_B=6p_C`

`[Ass um e(2/3)^0.4=0.85` and `R=(25)/(3)JK^-1mol^-1]`
Work done by the gas in the process `ArarrB` is

Options

A. (a) 5312J
B. (b) 1875J
C. (c) 6251J
D. (d) 8854J

Answer

B

Explanation

In adiabatic process,
`W_(AB)=-DeltaU_(AB)`
`=nC_V(T_A-T_B)`
`=(1)(3/2R)(T_A-T_B)`
`=(1)(3/2xx25/3)(1000-850)`
`=1875J`