Question

Two moles of an ideal gas originally at T = 400 K and V = 40 L undergo a free expansion;
?2 = 10?1
. Find (a) the entropy change of the gas, and (b) the entropy change of the universe?

Answer 1

It is isothermal expansion .

number of moles of ideal gas = 2 mol

Initial volume = 40 l

Final volume = 10 * initial volume.

= 400 L

Temperature = 400 k ( constant).

By First law of thermodynamics we have ,

At constant temperature change in internal energy = 0. ( also for free expansion of ideal gas, change in internal energy is zero.)

So,

Since it is expansion work is done by gas and bears - ve sign.

And For isothermal process , the work done by the gas.  

[BY DERIVATION IN THERMODYNAMICS PHYSICAL CHEMSTRY] -----------

So,

a) Entropy change , BY DEFINITION OF ENTROPY -

= 2.303* 2 * 8.314 j/mol.k * log40

= 61.35 j/k

Entropy change of the gas during expansion = 61.35 j/k

b)

Entropy change of the universe > 0 .Since there is no interaction between system ( ideal gas) and surroundings in free expansion of ideal gas .

Therefore ,

FOr free expansions of ideal gas.

So,

= 61.35 j/k + 0

So, it is said to be ,   for free expansions.

Note for reversible processes entropy change for universe in reversible processes = 0

Free expansions are spontaneous .

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W = Force * distance

F = Pressure *  Area

W = pressure * Area * distance

= pdV

    ( for ideal gas , pV = nRT)

[ Volume = area * length and length is movement of gas in cyllinder. So volume changes as movement of gas changes during expansion of the gas or contracion of the gas.]

W = pdV

net work done is determined by integration --

  

For isothermal processes --------

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