Question

Two moles of ideal He gas are contained at a pressure of 1 atm and a temperature of 300 K. 34166 J of heat are transferred to the gas, as a result of which the gas expands and does 1216 J of work against its surroundings. The process is reversible. (Note: C = 1.5R) Please calculate the final temperature of the gas

Answer 1

Explanation:

The given data is as follows.

       n = 2 mol,         P = 1 atm,         T = 300 K

        Q = +34166 J,         W= -1216 J (work done against surrounding)

       $C_{v}$ = $\frac{3R}{2}$

Relation between internal energy, work and heat is as follows.

      Change in internal energy ($\Delta U$) = Q + W

                                   = [34166 + (-1216)] J

                                   = 32950 J

Also,  $\Delta U = n \times C_{v} \times \Delta T$

                      = $3R \times (T_{2} - T_{1})$

                 32950 J = $3 \times 8.314 J/mol K \times (T_{2} - 300 K)$

                $\frac{32950}{24.942} = T_{2} - 300 K$

                            1321.06 K + 300 K = $T_{2}$    

                                       $T_{2}$ = 1621.06 K

Thus, we can conclude that the final temperature of the gas is 1621.06 K.