How Is Cp - Cv Calculated for an Ideal Monatomic Gas Using Thermodynamics?

[subzero0137]

Homework Statement

[/B]
Use the equation $$C_p - C_v = \left[ P + \left( \frac {∂U}{∂V} \right)_T\right] \left[ \left( \frac {∂V}{∂T} \right)_P \right]$$ to find $C_p - C_v$ for an ideal monatomic gas.

Homework Equations

$U = \frac {3}{2} RT$
$PV = RT$

The Attempt at a Solution

I substitute $PV = RT$ into the expression for $U$ to get $U = \frac {3}{2} PV$, therefore

$$\left( \frac {∂U}{∂V} \right)_T = \frac {3}{2} P$$

Since $PV=RT ⇒ V = \frac {RT}{P}$,

$$\left( \frac {∂V}{∂T} \right)_P = \frac {R}{P}$$

Therefore $C_p - C_v = [P + \frac {3}{2} P][\frac {R}{P}] = \frac {5}{2} R$ but the real answer is $C_p - C_v = R$. What have I done wrong? I can't seem to find the mistake.

 

[haruspex]

I substitute $PV = RT$ into the expression for $U$ to get $U = \frac {3}{2} PV$, therefore

$$\left( \frac {∂U}{∂V} \right)_T = \frac {3}{2} P$$

The T suffix in
$$\left( \frac {∂U}{∂V} \right)_T $$
means keeping T constant. In performing the differentiation you appear to have kept P constant instead.

 

[subzero0137]

The T suffix in
$$\left( \frac {∂U}{∂V} \right)_T $$
means keeping T constant. In performing the differentiation you appear to have kept P constant instead.

Oh I see. So the partial derivative would be 0?

 

[haruspex]

Oh I see. So the partial derivative would be 0?

Yes.