Question

A sample 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.

Answer 1

Explanation:

• For path A, the calculation will be as follows.

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

            W = $-2.303 nRT log(\frac{V_2}{V_1})$

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 ,

          $V_{2}$ = 8.34 L,    $V_{1}$ = 2.67 L

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

            W = $-2.303 nRT log(\frac{V_2}{V_1})$

   = $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

     = $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

    = -2818.68 J

Hence, work for path A is -2818.68 J.

• For path B, the calculation will be as follows.

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 $P_{external}$ = 1.00 atm.

So,              W = $-P_{external} \times \Delta V$

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

               W = $-P_{external} \times \Delta V$

                   = $-1 atm \times (8.34 L - 2.67 L)$  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

            = -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.