Derivation of work in expansion
Answers
Answer:
Derivation of work done in isothermal reversible expansion
dW=P×A×dl
=P×dV
The amount of work done by isothermal reversible expansion of an ideal gas from V
1
to V
2
is
W=∫
V
1
V
2
PdV
From ideal gas equation, P=
V
nRT
Hence W=∫
V
1
V
2
V
nRT
dV
=nRT∫
V
1
V
2
V
dV
On integration we get
W=nRTln
V
1
V
2
Since
P
1
V
1
=P
2
V
2
P
2
P
1
=
V
1
V
2
W=nRTln
P
2
P
1
Answer:
Solution:-
Let us consider n moles of ideal gas enclosed in a cylinder fitted with a weightless and frictionless piston. The work of expansion for a small change of volume dV against the external pressure P is given by-
dW=−PdV
∴ Total work done when the gas expands from initial volume V
1
to the final volume V
2
, will be-
W=−∫
V
1
V
2
PdV
For an ideal gas,
PV=nRT
⇒P=
V
nRT
Therefore,
W=−∫
V
1
V
2
V
nRT
dV
For isothermal expansion, T is constant.
Therefore,
W=−nRT∫
V
1
V
2
V
dV
⇒W=−nRT[lnV]
V
1
V
2
⇒W=−nRT(lnV
2
−lnV
1
)
⇒W=−nRTln(
V
1
V
2
)
⇒W=−2.303nRTlog(
V
1
V
2
).....(1)
At constant temperature,
P
1
V
1
=P
2
V
2
⇒
V
1
V
2
=
P
2
P
1
Therefore,
W=−2.303nRTlog(
P
2
P
1
).....(2)
Equation (1)&(2) are the expression for the work obtained in an isothermal reversible expansion of an ideal gas.