Answer:
Option B
Change in entropy of the process is [tex]\Delta S= Rln(\frac{P_{1}}{P_{2}})[/tex]
Explanation:
The entropy of a system is a measure of the degree of disorderliness of the system.
The entropy of a system moving from process 1 to 2 is given as
[tex]\Delta S = \int\limits^2_1 {\frac{\delta q}{T}}[/tex]
recall from first law, [tex]\delta q =du +Pdv[/tex]
hence we have, [tex]\Delta S = \int\limits^2_1 {\frac{du +Pdv}{T}}[/tex]
since the process is isothermal, [tex]du= 0[/tex]
this gives us [tex]\Delta S = \int\limits^2_1 {\frac{Pdv}{T}}[/tex]
integrating within the limits of 1 and 2, will give us
[tex]\Delta S = R ln (\frac{V_{2}}{V_{1}})[/tex]
also from ideal gas laws,
[tex]\frac{V_{2}}{V_{1}}=\frac{P_{1}}{P_{2}}[/tex]
hence we have [tex]\Delta S = R ln (\frac{P_{1}}{P_{2}})[/tex]
This makes the correct option B
The correct answer is (d). R ln(P1/P2)
In Isothermal Process, the heat exchange takes place at constant temperature ΔT=0
dQ = dU + dW
Here dU = 0 since there is no internal energy change at constant temperature.
dW = PdV
but here pressure is changing,
PV = nRT, given n= 1 ⇒ PV = RT
V = RT/P ⇒ dV = -RT/[tex]P^{2}[/tex] dP
⇒ dQ = -RT/P dP
So the change in entropy,
ΔS = ∫dQ/T = [tex]-\int\limits^{P_{2}}_{P_{1}} {\frac{R}{P} } \, dP[/tex] = R ln([tex]P_{1}[/tex]/[tex]P_{2}[/tex])
Learn more about Entropy:
https://brainly.com/question/104730
