A thermodynamic system is taken through the cycle abcda. (a) calculate the work done by the gas during the parts ab, bc, cd and da. (b) find the total heat rejected by the gas during the process.

Explanation

`n=(1)/(2)mol.`,
`R=25/3J//mol.*k`
`gamma=5/3`
(a) Temperature at `a=Ta`,
`PaVa =nRTa`
implies `Ta= (PaVa)/(nR)=120 K`
Similarly, temperature at `b=240K`, at c it is `480K` and at `d` it is `240K`.
(b) For `ab` process
`dQ =nc_pdT`
[Since `ab` is isometric]
`=1/2xx (Rgamma)/(gamma-1)(Tb-Ta)`
`=1/2xx ((25xx5)/(3xx3))/(5/3-1)xx(240-120)`
`=1/2xx (125)/(9)xx(3)/(2) xx (120)`
`1250J`.
For `bc`, `dQ = dU + dW`
[`dW =0`, Isochoric process]
`=nCv(Tc-Tb)`
`=1/2 xx (((25)/(3)))/([(5/3)-1])xx(240)`
`=1/2xx25/3xx3/2xx240=1500J`.
(c) Heat liberated in `cd =-nC_pdT`
`=-1/2xx (gammaR)/(gamma-1) xx ((Td)/(Tc))`
`=-1/2xx125/9xx3/2 xx(240-480)`
`=-1/2xx 125/6xx240=2500J`
Heat liberated in `da`
`=-nCvdT`
`=-1/2 xx (R)/(gamma-1)(Ta-Td)`
`=-1/2 xx25/2 xx (120-240)`
`=25/4xx120 = 750J`.