One mole of an ideal monatomic gas udnergoes thermodynamic cycle `1 rarr 2 rarr 3 rarr 1` as shown in figure. Initial temperature of gas in `T_(0)=300K`.
Process ` 1 rarr2 : P=a V`
Process ` 2 rarr 3: PV=` Constant
Process `3 rarr 1 : P=` Constant
`(` Take `1n |3|=1.09)`

Find the net work done by the cycle.

Options

A. `3.27 RT_(0)`
B. `6.83 RT _(0)`
C. `4.53 RT_(0)`
D. `5.81 RT _(0)`

Answer

D

Explanation

For process `1 rarr 2 `
`W_(12)=int_(1)^(2)alphaVdV=alphaint _(V_(0))^(3V_(0))VdV=(alpha)/(2)( 9 V_(0)^(2)-V_(0)^(2))`
`=4 alpha V_(0)^(2)`
Using gas law, `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
`T_(2)=(P_(2)V_(2))/(P_(1)V_(1))T_(1)=(V_(2)^(2))/(V_(1)^(2))T_(1)=((3V_(0))/(V_(0)))^(2)T_(0)=9T_(0)`
For process `2 rarr 3`
`W_(23)=RT_(2)log|(P_(2))/(p_(3))|=R(9T_(0))log |(3P_(0))/(P_(0))|`
`=9RT_(0)log|3|=9.81 RT_(0)`
For isothermal process`: P_(2)V_(2)=P_(3)V_(3)`
Therefore, `V_(3)=(P_(2)V_(2))/(P_(3))=(3P_(0))/(P_(0))(3V_(0))=9V_(0)`
Also, `W_(31)=P_(0)(V_(1)-V_(3))=P_(0)(V_(0)-9V_(0))=-8 P_(0)V_(0)`
`=-8 RT_(0)`
Applying gas law in process `1 rarr 2`
`P_(0)V_(0)=RT_(0)` or` alphaV_(0)^(2)=RT_(0)`
The net work is
`W_(n et)=W_(12)+W_(23)+W_(31)`
`=4RT_(0)+9.81RT_(0)-8RT_(0)` ltbr. `=5.81 RT_(0)`
For process `1 rarr 2`
`DeltaU_(12)=C_(V)(T_(2)-T_(1))`
`=4 RT_(0)+12 RT_(0)=16 RT_(0)`
Since
`DeltaU_(12)=C_(12)(T_(2)-T_(1))=8 C _(12)T_(0)`
` C_(12)=2 R=16.6 J // mol -K`
For the process `2 rarr 3: C_(23) = oo`
For the process ` 3 rarr 1 : C _(31)=C_(P)+R`
`=5 R //2 =20/75 J//mol-K`