Question

Two identical thermally insulated spherical tanks, A and B, are connected by a valve. Initially tank A contains 20mol of an ideal diatomic gas, tank B is evacuated, and the valve is closed.
If the valve is opened and the gas expands isothermally fromA to B, what is the change in the entropy of the gas?

Answer 1

Answer:

Explanation:

Volume of gas is doubled isothermally . Let the temperature of the gas be T .Work done by the gas

W = 2.303 nRT log 2 V / V

Since expansion is isothermal , Δ E = 0

Q = ΔE + W

Q = W

Q = 2.303 nRT log 2 V / V

Q / T = 2.303 n R log 2

Change in entropy = ΔS = Q / T = 2.303 n R log 2

= 2.303 x 20 x 8.3 x .3010 J T⁻¹.

= 115  J T⁻¹.

Answer 2

The required change in the entropy of the gas is 115 J/K.

Given data:

The number of moles of diatomic gas in tank A is, n = 20 mol.

The expansion takes place isothermally, which means the volume is doubled at the constant temperature. Therefore, work done during the isothermal expansion is,

W = 2.303 nRT log 2 V / V

Since expansion is isothermal , Δ E = 0. Then by using the first law of thermodynamics as,

Q = ΔE + W

Q = W

Q = 2.303 nRT log 2 V / V

Q / T = 2.303 n R log 2

Also, we that the expression for the change in Entropy is,

 ΔS = Q / T = 2.303 n R log 2

Solving as,

ΔS = 2.303 x 20 x 8.3 x .3010

ΔS = 115  J/K

Thus, we can conclude that the required change in the entropy of the gas is 115 J/K.

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