Question

A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure P1 to pressure P2 while losing an amount q of heat to the surroundings at temperature T. If the gas constant of the gas is R, the entropy change of the gas âs during this process is:a. âS =R ln (P2/P1)b. âS = R ln (P1/P2)c. âS =R ln (P2/P1) - q/Td. âS = R ln (P1/P2) - q/Te. 0

Answer 1

Answer:

Option (b) is the correct answer

$\Delta S = RIn(\frac{P_1}{P_2} )$

Explanation:

Given

$T_1=T_2=T$  = constant

we know for ideal gas ,

U = F(T)

Therefore,

T = constant ⇒ U = constant

⇒ΔU = 0

or ΔU = CV(T₂ - T₁) ⇒ Δ U = 0

we know,

Pv = RT

⇒ Pv = constant  = C

work done(specific)

$\Delta w = \int\limits {Pdv} = \frac{c}{v} \int\limits^{v_2}_{v_1} {} \, dv$

$= c In (\frac{v_2}{v_1} )$

⇒Δw = P₁v₁ In(v₂/v₁)

Also,

Pv = constant

⇒P₁v₁=P₂v₂

⇒v₂/v₁ = P₁/P₂

Δw = P₁v₁ In(P₁/P₂)

By the first law of thermodynamic

Δq = Δw + Δu

⇒Δq = P₁v₁In(P₁/P₂) + 0

Now,

Δs = Δq / T

= $\frac{P_1v_1}{T_1} In\frac{P_1}{P_2}$

$\Delta S = RIn(\frac{P_1}{P_2} )$

Answer 2

Answer:

Option B

Change in entropy of the process is $\Delta S= Rln(\frac{P_{1}}{P_{2}})$

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

$\Delta S = \int\limits^2_1 {\frac{\delta q}{T}}$

recall from first law, $\delta q =du +Pdv$

hence we have, $\Delta S = \int\limits^2_1 {\frac{du +Pdv}{T}}$

since the process is isothermal, $du= 0$

this gives us $\Delta S = \int\limits^2_1 {\frac{Pdv}{T}}$

integrating within the limits of 1 and 2, will give us

$\Delta S = R ln (\frac{V_{2}}{V_{1}})$

also from ideal gas laws,

$\frac{V_{2}}{V_{1}}=\frac{P_{1}}{P_{2}}$

hence we have    $\Delta S = R ln (\frac{P_{1}}{P_{2}})$

This makes the correct option B