Question

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothermal process? kJ (b) How much work is required to produce the same compression in an adiabatic process? kJ (c) What is the final pressure in part (a)? atm (d) What is the final pressure in part (b)? atm

Answer 1

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

        W = $-nRT ln(\frac{V_{2}}{V_{1}})$

             = $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

             = $-12280.82 \times ln (0.09)$

             = $-12280.82 \times -2.41$

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

              = $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

              = $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               $P_{1}V_{1} = P_{2}V_{2}$

or,       $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

                    = $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          $P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or,       $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

                    = $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.