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, ( c) the amount of heat supplied in the path ab and in the path bc and (d) the change in the internal energy of the gas 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`.