Question

(a) Two moles of an ideal gas occupy a volume V. The gas expands isothermally and reversibly to a volume 3V. (i) Is the velocity distribution changed by the isothermal expansion? Explain. (ii) Use Eq. (Microscopic state) to calculate the change in entropy of the gas. (iii) Use Eq. (reversible isothermal process) to calculate the change in entropy of the gas. Compare this result to that obtained in part (ii).

Answer 1

Gives

n=2 mol

$V_1=V$

$V_2=3V$

(a)

In the isothermal process the temperature T is constant.

As we know that molecules in a gas don't all have the same speed also this speed depends on the temperature T therefore the temperature is constant or change the distribution velocities of the molecules would not change. where the distribution function depends on the temperature T not on the change in temperature ∆T.

Part (b) entropy change

∆S =$kln(\frac{w_2}{w_1})\rightarrow(1)$

$w_2=3^Nw_1$

Put value

Stage w₁ the gas has number of molecule (N )

N=nN_A$\rightarrow(2)$

∆S=$kln\frac{3^{N}w_1}{w_2}$

∆S=$Nkln3\rightarrow(3)$

equation (2)and(3)we can written as

∆S=$nN_Akln3$

Where

$K=1.38\times10^{-23}J/K$

$$N_A=6.023\times10^{23} molecule/mol$$

Put values

∆S

$$=2\times6.023\times10^{23}\times1.38\times10^{-23}ln3$$

Where

$ln3=1.0986$

∆S=18.26J/k

Part(c)

Change of entropy

∆S=$\frac{Q}{T}\rightarrow(5)$

$Q=W+∆U$

Isothermal process

∆U=0

Q=W

$Q=W=nRT ln\frac{V_2}{V_1}\rightarrow(6)$

$Q=W=nRTln\frac{3V}{V}$

$Q=W=nRTln3$

∆S=$\frac{Q}{T}\rightarrow(7)$

∆S=$\frac{nRTln3}{T}$

S=nR$ln3$

Put values

∆S$=2\times8.314\times ln3J/K$

∆S=18.26J/k

Solution

Part(a)

No,the distribution velocity does not change

Part(b)

∆S=18.26J/k

Part(c)

∆S=18.26J/k

The same for part (b)