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?

Explanation:

The given data is as follows.

       n = 1 mol,    

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

(a)    

And, according to the first law of thermodynamics

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 =

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

            Q =

Now,  

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

                      =

                       = nR ln 2

                        =

                        = 5.76 J/K

Therefore, change in entropy is 5.76 J/K.

(b)    As,  Q =

                   =

                   = nRT ln 2

           T =

              =

              = 260.4 K

Therefore, temperature of the gas is 260.4 K.

if you're asking what comes out, it's steam.