For a given ideal gas `6 xx 10^(5)` J heat energy is supplied and the volume of gas is increased from 4 `m^(3)` to `6 m^(3)` at atmospheric pressure . Calculate (a) the work done by the gas ( b) Change in internal energy of the gas
( c) graph this process in PV and TV diagram

Explanation

Mayer.s relation: Consider u mole of an ideal gas in a container with volume V, pressure P and temperature T.
When the gas is heated at constant volume the temperature increases by dt. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU.
If `C_v` is the molar specific heat capacity at constant volume, from equation.
` C_v = 1/mu ( d U )/(dT )`
`dU = mu C _v dT`
Suppose the gas is heated at constant pressure so that the temperature increases by dT. If .Q. is the heat supplied in this process and .dV. the change in volume of the gas.
` Q= mu C _(p ) dT `...(3)
If W is the work done by the gas in this process, then
`W= PDV`
But from the first law of thermodynamics,
`Q=dU + W `
Substituting equations (2), (3) and (4) in (5), we get,
`mu C _(p) dT = mu C_(v) dT + p dV `
For mole of ideal gas, the equation of state is given by
`PV = mu RT implies pd V + dp = mu R d T`
Since the pressure is constant, dP = 0
`therefore C_P dT = C_(v) dT + RdT `
` therefore C_(P) = C_(V) +R (or ) C_(P ) - C_(V ) = R `
This relation is called Mayer.s relation It implies that the molar specific heat capacity of an ideal gas at constant pressure is greater than molar specific heat capacity at constant volume
the realation shows that specific heat at constant pressure `(s_p)` is always than calculate the work done by the gas
(ii) A gas expands from volume `1m^3` to `2m^3` at constant atmospheric pressure.
(ii ) The pressure P = 1 atm = 101 kPa,
` V_f = 2 m^3 and V_i = 1 m^3`
From equation ` W = int_(v_i)^(v_f) pd V = Pint _(vi)^(vi) dv `
Since P is constant. It is taken o’t of the integral.
`W= P (V_f -V_f ) = 10 1 xx 10^3 xx ( 2-1) = 101 KJ`