Question

Question No. 13
A А
A
A*
Entropy change for an ideal gas when process is
isothermal (T1 =T2)
AS = nRIn(P1
/P2)
AS = Cp Inc
T2/T)
AS = CV
In(T2/T1)
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Answer 1

Answer:

∆S=nRln(P1/P2)

Explanation:

From first law of thermodynamics

∆U=∆Q+∆W

∆U will be zero because it is an isothermal process.

So,

∆Q= -∆W

∆Q= -[-nRTln(P1/P2)]

∆Q= nRTln(P1/P2)

As we know that

∆S= ∆Q/T

Putting the value of ∆Q

we will get the answer as

∆S=nRln(P1/P2)

hope it helps....

Answer 2

$\delta S=nR*ln(\frac{P_{1} }{P_{2} } )$

       

Explanation:

We have

$\Delta U=\Delta Q+\Delta W$

∆U = 0 in an isothermal process.

So,

$\Delta U=-\Delta W$

$\Delta Q= nRT*ln\frac{P_{1}} {P_{2}}$

Now

$\Delta S= \frac{\Delta Q}{T}$

Putting the value of ∆Q we get

$\Delta S=nR*ln\frac{P_1}{P_2}$