# Work done in adiabatic process vs work done in isothermal

## 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##

## 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 ?

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