- #1

shihab-kol

- 119

- 8

## 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)

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 ?

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 ?