Work done in adiabatic process vs work done in isothermal

[shihab-kol]

Homework Statement

[/B]
An ideal gas is compressed to the same volume from the same initial state for both an adiabatic and an isothermal process. In which case will more work be done ?

2. Homework Equations

$dU=dQ - dW$
$W=\int P\,dV$(For isothermal)
$W=nc_vdT$
$W=\frac{(P_1V_1-P_2V_2)} {ϒ-1}$(For adiabatic)
$PV^ϒ=constant$(For adiabatic)

The Attempt at a Solution

I take the initial state to be $(P_1,V_1)$ and the final as $(P_2,V_2)$ for isothermal and $(P_3,V_2)$ for adiabatic.
I have $$PV=nRT$$
Differentiating with respect to V,
(For isothermal)[/B]
$$\frac{dP_i}{dV_i}=-\frac{P}{V}$$
And,
(For adiabatic)$$\frac{d(PV^ϒ)}{dV}= 0$$
Or,$$\frac{dP_a}{dV_a}=-(ϒ-1)\frac{P}{V}$$
Substituting the earlier equation into the above,$$\frac{dP_a}{dV_a} = (ϒ-1)\frac{dP_i}{dV_i}$$
What I get is that the slope of an adiabat is higher than the slope of an isothermal having the same initial state and thus,
$$P_3>P_2$$
And,
$$W_{Adiabatic}>W_{Isothermal}$$
Is this alright ?

 

[Chestermiller]

Your answer is correct.

 

[shihab-kol]

Your answer is correct.

Thanks for taking the trouble of checking it out.