One mole of a monoatomic ideal gas is taken through the cycle shown in figure.
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`ArarrB` Adiabatic expansion
`BrarrC` Cooling at constant volume
`CrarrD` Adiabatic compression.
`DrarrA` Heating at constant volume
The pressure and temperature at A,B etc., are denoted by `p_A, T_A, p_B, T_B` etc. respectively.
Given, `T_A=1000K`, `p_B=(2/3)p_A` and `p_C=(1/3)p_A`. Calculate
(a) the work done by the gas in the process `ArarrB`
(b) the heat lost by the gas in the process `BrarrC`
Given, `(2/3)^0.4=0.85` and `R=8.31J//mol-K`

Answer

A, B

Explanation

(a) As for adiabatic change `pV^gamma=` constant
i.e. `p((nRT)/(p))^gamma=` constant (as `pV=nRT`)
i.e. `(T^gamma)/(p^(gamma-1))=` constant so`((T_B)/(T_A))^gamma=((p_B)/(p_A))^(gamma-1)`, where `gamma=5/3`
i.e. `T_B=T_A(2/3)^(1-1/gamma)=1000(2/3)^(2//5)=850K`
So, `W_(AB)=(nR[T_F-T_I])/([1-gamma])=(1xx8.31[1000-850])/([(5/3)-1])`
i.e. `W_(AB)=1869.75J`
For `BrarrC`, V=constant so `DeltaW=0`
So, from first law of thermodynamics
`DeltaQ=DeltaU+DeltaW=nC_VDeltaT+0`
or `DeltaQ=1xx(3/2R)(T_C-850)` (as `C_V=3/2R` )
Now, along path BC, V=constant, `p porp T`
i.e. `(p_C)/(p_B)=(T_C)/(T_B)`
`T_C=((1/3)p_A)/((2/3)p_A)xxT_B=T_B/2=850/2=425K` ...(ii)
So, `DeltaQ=1xx3/2xx8.31(425-850)=-5297.625J`
[Negative heat means, heat is lost by the system]