Half mole of an ideal gas `(gamma = 5/3) is taken through the cycle abcda as shown in figure . Take `(R = 25/3 JK^(-1) mol^(-1))`. (a) Find the temperature of the gas in the staes a, b, c and d. (b) Find the amount of heat supplied in the processes ab and bc. (c ) Find the amount of heat liberated in the processes `cd and da`.

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`.