Question

cp-cv=R? for ideal gas

Answer 1

Question 36330

Let us first write the first law of thermodynamics $\delta Q = dU + pdV$ .

For constant pressure, $\delta Q = dU$ . Hence, heat capacity for constant pressure is

$$C _ {V} = \left(\frac {\delta Q}{d T}\right) _ {V} = \left(\frac {\partial U}{\partial T}\right) _ {V}.$$

For an ideal gas, internal energy $U$ is the function only of temperature $(U = U(T))$ . Hence,

$$C _ {V} = \frac {d U}{d T}.$$

Now, by definition

$$C _ {P} = \left(\frac {\delta Q}{d T}\right) _ {P} = \left(\frac {d U + p d V}{d T}\right) _ {P} = \left(\frac {\partial U}{\partial T}\right) _ {P} + P \left(\frac {\partial V}{\partial T}\right) _ {P}.$$

For one mole of ideal gas, equation of state is $PV = RT$ , which yields $\left(\frac{\partial V}{\partial T}\right)_P = \frac{R}{P}$ and

$\left(\frac{\partial U}{\partial T}\right)_P = \frac{dU}{dT} = C_V$ . Plugging in last two equalities into formula for $C_P$ , obtain

$C_P = C_V + R$ - this is Mayer's formula.

This formula works only for ideal gas, because we have used the equation $PV = RT$ for ideal gas, while deriving it.