Answer to Question #193537 in Molecular Physics | Thermodynamics for Taamba Shikoko

Question

Two moles of an ideal gas occupy a volume V. The gas expands isothermally and reversibly to a volume

3V. (a) Is the velocity distribution changed by the isothermal expansion? Explain. (b) Use Eq.

(Microscopic state) to calculate the change in entropy of the gas. (c) Use Eq. (reversible isothermal

process) to calculate the change in entropy of the gas. Compare this result to that obtained in part (b).

Answer 1

Given

We are given an ideal gas with a number of moles n = 2.0 mol expands

isothermally from V₁ to V₂= 3V₁.

Solution

(a) In the isothermal process, the temperature T is constant. As we know , the 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 $\Delta T$ .

(b)

We are asked to calculate the change in entropy $\Delta S$

which is expressed by

$\Delta S=kln\frac{w_2}{w_1}$ .............(1)

Where k is Boltzmann's constant and equals$1.38x 10^{-23} J/K$

w₁ is the number of possible microscopic states for a given macroscopic

intial state and w₂ is the same for the final state.

At the intial state w₁, the gas has number of molecules N given by

$N = nN_A$

Where N_A is Avogadro's number and equals 6.023x 10²³molecules/mol

When the gas expands to three times volume 3V₁ the number

of microscopic final state is greater than at initial state by factor 3^N, so w₂

will be

$w_2 = 3^Nw_1$

Now substitute this value of w₂ into equation (1)

$\Delta S = N k In 3 \ \ \ \ \ \ \ \ \ \ \ \ (N = nN_A) \\\Delta S = n N_A k In 3 \ \ \ ..........(2)$

Now we can plug our values for N_A, n and k into equation (2)

= 2.0 mol x 6.023 x 1023 molecules/mol x 1.38 x 10-23J/K x In 3

18.10J/K

=$\boxed{18.10 J/k}$

(c)

We are asked to calculate the change in entropy $\Delta S$

$\Delta S= \frac{Q}{T}$ .................(3)

Where Q is the heat gained by the gas and could be calculated by the first

law of thermodynamics Q = W + $\Delta U$ where $\Delta U$ equals zero as the process

is isothermal. Therefore Q will be

$Q = W = nRT In\frac{V_2}{ V_1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (V_2=3V_1)\\ Q = W = nRT In\frac{3V_1}{ V_1} \\ Q = W = nRT In3$

Where R is the gas constant and equals 8.314 J/mol.K

Now plug this equation for Q into equation (3) to get $\Delta S$

$\Delta S = nRT In3=2mole\times8.314J/mol.k\times ln3\\ \Delta S= \boxed{18.10 J/k}$

The change in entropy is the same for part (b)