derive the equation Wmax=-2.303 nrt log10 v2/v1
Answers
Answer:
Solution:- (C) 2.303nRT×log(
V
2
V
1
)
Consider 'n' moles of an ideal gas enclosed in a cylinder fitted with a weightless, frictionless, airtight movable position. Let the pressure of the gas be P which is equal to external atmospheric pressure P. Let the external pressure be reduced by an infinitely small amount dP and the corresponding small increase in volume be dV.
Therefore the small work done in the expnsion process will be-
dW=−P
ext.
dV
⇒dW=−(P−dP)dV
⇒dW=−P.dV+dP.dV
Since both dP and dV are very small, the product dP.dV will be very small in comparison with P.dV and thus can be neglected.
∴dW=−P.dV
When the expansion of the gas is carried out reversibly then there will be a series of such P.dV terms. Thus the total maximum work W
max
can be obtained by integrating this equation between the limits V
1
to V
2
. Where V
1
is initial volume while V
2
is final value.
W=∫
V
1
V
2
dW
W=∫
V
1
V
2
(−P.dV)
As we know that, from ideal gas equation,
PV=nRT
⇒P=
V
nRT
W=−∫
V
1
V
2
V
nRT
dV
W=−nRT∫
V
1
V
2
V
dV
W=−nRT[lnV]
V
1
V
2
W=−nRT(lnV
2
−lnV
2
)
W=nRT(lnV
1
−lnV
2
)
W=nRTln(
V
2
V
1
)
W=nRT(2.303×log(
V
2
V
1
))
W=2.303nRT×log(
V
2
V
1
)
Hence work done in reversible isothermal process by an ideal gas is given by-
W=2.303nRT×log(
V
2
V
1
)
Explanation:
Reversible Isothermal Process.
