Question

Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected by a tube that contains an ideal turbine (a mechanical device) and a valve. Initially, the valve is closed. Container A (volume VA = 2 m3) is filled with n = 1.2 moles of an ideal monatomic gas, and container B (volume VB = 3.5 m3) is empty (vacuum). When the valve is opened, gas flows between A and B, and the turbine is used to generate electricity. What is the maximum amount of work than can be done on the turbine as the gas reaches equilibrium? Remember: U = 1.5 nRT + const S = nR ln(V) + f(U,n)

Answer 1

Answer:

The maximum amount of work is  $W = 1563.289 \ J$

Explanation:

From  the question we are told that

   The temperature of the environment is  $T = 280\ K$

    The volume of container A is  $V_A = 2 m^3$

    Initially the number of moles  is  $n = 1.2 \ moles$

     The volume of container B is $V_B = 3.5 \ m^3$

     

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

             $W = P_A V_A ln[ \frac{V_B}{V_A} ]$

Now from the Ideal gas law

          $P_A V_A = nRT$

So substituting for $P_A V_A$ in the equation above

          $W = nRT ln [\frac{V_B}{V_A} ]$

Where R is the gas constant with a values of  $R = 8.314 \ J/mol$

Substituting values we have that

            $W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]$

          $W = 1563.289 \ J$