Derive Mayer's relation for an ideal gas.

Explanation

Consider `mu` mole of an ideal gas enclosed in a cylinder provided with a frictionless piston of area A.
P - pressure of gas, V - volume of gas
T - absolute temperature gas
dQ - quantity of heat supplied
To keep the volume of the gas constant a small `w_(1)` is placed over the piston.
The pressure and temperature increase to `P+dP` and `T+dT`.
dQ is used to increase the internal energy dU of the gas. But the gas does not do any work `[dW=0]`
`therefore dQ=dU=1xxc_(v)xxdT`.
Now the `w_(t)` is removed. The piston now moves upwards thus a distance dx, the pressure of the encloed gas equal to atmosphere pressure P. Due to expansion, temperature decreases.
Now the `w_(t)` is removed. The piston now moves upwards thus a distance dx, the pressure of the enclosed gas equal to atmosphere pressure P. Due to expansion, temperature decreases.
Now a quantity of heat dQ' is supplied till its temperature become `T+DeltaT`. This heat energy is not only used to increase the internal energy dU of the gas but also to do exists work dW in moving the piston upwards.
`therefore dQ'=dU+dW`
At constant pressure `dQ^(1)=c_(p)dT`
`therefore c_(p)dT=c_(v)dT+dW`
Work done dW = Force `xx` distance = `PxxAxxdx`
dW = P.dV [A.dx = dV (change volume)]
`therefore c_(p)dT=c_(v)dT+PdV" " (1)`
The equation of state of an ideal gas is
`PV=RT`
Differentiate on both the sides
`PdV=RdT" "(2)`
Substitute (2) in (1)
`c_(p)dT=c_(v)dT+RdT`
`c_(p)=c_(v)+R`
`therefore c_(p)-c_(v)=R`
This equation is known as Meyer's electron.