An ideal gas `(gamma = 1.67 )` is taken through the process abc shown in figure . The temperature at the point a is `300 k`. Calculate (a) the temperature at b and c, (b) the work done in the process.

Explanation

(a) For a, b 'V' is constant
So, `(P_1)/(T_1) = (P_2)/(T_2)`
implies `(100)/(300) = (200)/T_2`
`T_2=((200xx300)/(100))=600K`
For b, c 'p' is constant
So, `V_1/T_1 = V_2/T_2`
implies `100/600 = 150/T_2`
implies `T_2((600xx150)/(100)) = 900K`
(b) Work done = Area enclosed under the graph `50cc xx 20 KPa`
`=50xx10^(-6)xx200xx10^(3) J`
`=10J`.
(c) 'Q' supplied `= nC_vdT`
Now,
`Qbc =((PV)/(RT))xx((R)/(gamma-1)) xxdT`
`=(200xx10^(3) xx 100 xx 10^(-6) xx 300)/ (600xx0.67)
`=14.925` `( :. gamma = 1.67)`
'Q' supplied to `bc = nCpdT` (`C_p=(gammaR)/(gamma-1)`)
`=(PV)/(RT) xx (gammaR)/(gamma-1) xxdT`
`=(200xx10^(3) xx 150 xx 10^(-6))/(600xx0.67)xx 300`
`=10xx(1.67)/(0.67)=24.925`
(d) `Q= DeltaU+W`
Now,
`DeltaU=Q-W`
=Heat supplied-Work done
`=(24.925 + 14. 925)-10`
`=39.850-10=29.850J`.