Five moles of an ideal gas are taken in a Carnot engine working between `100^circC` and `30^circC`. The useful work done in one cycle is 420 joule. Calculate the ratio the volume of the gas at the end beginning of the isothermal expansion. `R=8.4 J Mole^-1 K^-1`.

Answer

`1.153`

Explanation

Here, `n=5, T_(1)= 100^(@)C= (100+273)K = 373K`,
`T_(2)= 30^(@)C=(30+273)K=303K`,
`W=420J, (V_(2))/(V_(1))=?, R= 8.4J mol e^(-1)K^(-1)`
As `(Q_(1))/(Q_(2))=(T_(1))/(T_(2))=(373)/(303) :. Q_(1)(373)/(303)Q_(2)`....(i)
Now, `W=Q_(1)-Q_(2)=(373)/(303)Q_(2)-Q_(2)=(70)/(303)Q_(2)`
`:. Q_(2)=(373)/(70)W= (303)/(70)xx420=1818J`
From (i), `Q_(1)=(373)/(303)xx1818= 2238J`
Work done during isothermal expansion
`W_(1)=Q_(1)= 2.3026 n RT_(1)"log"_(10)(V_(2))/(V_(1))`
`2238= 2.3026xx5xx8.4xx373 "log"_(10)(V_(2))/(V_(1))`
`"log"_(10)(V_(2))/(V_(1))=(2238)/(2.3026xx5xx8.4xx373)=0.0620`
`(V_(2))/(V_(1))= antilog0.0620= 1.153`