Question

One mole of an ideal gas doubles its volume in a reversible isothermal expansion. (a) What is the change in entropy of the gas? (b) If 1500 J of heat are added in this process, what is the temperature of the gas?

Answer 1

Explanation:

The given data is as follows.

       n = 1 mol,     $V_{f} = 2V_{i}$

       Q = 1500 J,      R = 8.314 J/mol k

(a)    $\Delta S = \frac{dQ}{dT}$

And, according to the first law of thermodynamics

                $\Delta E_{int} = Q - W$

And, in an isothermal process the change in internal energy of the gas is zero.

Hence,    0 = Q - W

or,             W = Q

Expression for work done in an isothermal process is as follows.

                   W = $nRT ln \frac{V_{f}}{V_{i}}$

As W = Q, Hence expression for Q will also be given as follows.

            Q = $nRT ln \frac{V_{f}}{V_{i}}$

Now,  

        $\Delta S = \frac{nRT ln \frac{V_{f}}{V_{i}}}{T}$

        [/tex]\Delta S = nR ln \frac{V_{f}}{V_{i}}[/tex]

                      = $nR ln \frac{2V_{i}}{V_{i}}$

                       = nR ln 2

                        = $1 \times 8.314 \times 0.693$

                        = 5.76 J/K

Therefore, change in entropy is 5.76 J/K.

(b)    As,  Q = $nRT ln \frac{V_{f}}{V_{i}}$

                   = $nRT ln \frac{2V_{i}}{V_{i}}$

                   = nRT ln 2

           T = $\frac{Q}{nR ln 2}$

              = $\frac{1500}{1 \times 8.314 ln 2}$

              = 260.4 K

Therefore, temperature of the gas is 260.4 K.