Asample of an ideal gas in a cylinder of volume 2.67 l at 298 k and 2.81 atm expands to 8.34 l by two different pathways. path a is an isothermal, reversible expansion. path b has two steps. in the first step, the gas is cooled at constant volume to 1.00 atm. in the second step, the gas is heated and allowed to expand against a constant external pressure of 1.00 atm until the final volume is 8.34 l. calculate the work for path a and b.

Explanation:

As, for reversible isothermal expansion the formula is as follows.

            W =

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          = 8.34 L,     = 2.67 L

Now, putting the given values into the above formula as follows.

            W =

   =

     =

    = -2818.68 J

Hence, work for path A is -2818.68 J.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure = 1.00 atm.

So,              W =

Now, putting the given values into the above formula as follows.

               W =

                   =  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W =

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

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