Work done in adiabatic process vs work done in isothermal

  • #1
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)


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 ?

 

Answers and Replies

  • #3
119
8
Your answer is correct.
Thanks for taking the trouble of checking it out.
 

Related Threads on Work done in adiabatic process vs work done in isothermal

Replies
10
Views
72K
Replies
2
Views
15K
Replies
1
Views
893
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
8
Views
24K
Replies
4
Views
3K
Replies
5
Views
3K
Replies
2
Views
3K
  • Last Post
Replies
4
Views
781
  • Last Post
Replies
3
Views
1K
Top