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

1.15

Explanation

Here `T_1 = 100+ 273 = 373K`,
`T_2 = 30 +273 = 303 K`
Amount of ideal gas = 5 mole
Useful work done in one cycle,
`W = Q_1-Q_2 = 420J`
Now, `Q_1/Q_2 - T_1/T_2 =(373)/303 ` or `Q_1 = 373/303 Q_2 " "...(ii)`
From equations (i) and (ii), we have
`373/303 Q_2-Q_2 = 420 ` or `70/303 Q_2=420`
or `Q_2 = (420xx303)/70 = 1818 J`
From equation (i),
`Q_1 = Q_2 +420 = 1818 +420 = 2238 J.`
When the gas is carried through a Car not cycle, the heat absorbed `Q_1` during the isothermal expansion is equal to the work done by the gas. If `V_1 and V_2` are the volumes of the gas at the beginning and at the end of the isothermal expansion, then `Q_1 = 2.303 nRT_(1) log_(10)""V_2/V_1`
or `2238 = 2.303 xx5xx8.4xx373 log_(10)""V_2/V_1`
`log_(10)""V_2/V_1=2238/(2.303 xx5 xx8.4 xx373)=0.0620`
Hence `V_2/V_1 = 1.15`