Question

Carbon dioxide (CO2) contained in a piston–cylinder arrangement, initially at 6 bar and 400 K, undergoes an expansion to a final temperature of 298 K, during which the pressure–volume relationship is pV1.2 = constant. Assuming the ideal gas model for the CO2, determine the final pressure, in bar, and the work and heat transfer, each in kJ/kg.

Answer 1

Explanation:

The given data is as follows.

          Initial temperature ($T_{1}$) = 400 K

          Final temperature ($T_{2}$) = 298 K

            Process $P \nu^{1.2}$ = C

For $CO_{2}$, according to the properties of various substances the values are as follows.

          R = 0.1889 kJ/kg K

       $C_{p}$ = 0.846 kJ/kg K

Therefore, calculate the final pressure as follows.

         $\frac{P_{2}}{P_{1}} = (\frac{T_{2}}{T_{1}})^{\frac{n}{n-1}}$

         $\frac{P_{2}}{6} = (\frac{298}{400})^{\frac{1.2}{1.2-1}}$

            $P_{2}$ = 1.025 bar

Hence, final pressure is 1.025 bar.

Now, work done will be calculated as follows.

             W = $\frac{P_{1}\nu_{1} - P_{2}\nu_{2}}{n - 1}$

                 = $\frac{m(RT_{1} - RT_{2})}{n - 1}$

                 = $\frac{1 \times 0.1889(400 - 298)}{(1.2 - 1)}$

                 = 96.339 kJ/kg

Hence, work done is equal to 96.339 kJ/kg.

Formula to calculate the change in internal energy is as follows.

             $\Delta U = C_{v}(T_{2} - T_{1})$

                         = 0.657 (298 - 400)

                         = -67.014 kJ/kg

Now, transfer of heat will be calculated as follows.

                 Q = $\Delta U + W$

                     = -67.014 kJ/kg + 96.339 kJ/kg

                     = 29.325 kJ/kg

Therefore, amount of heat transfer is 29.325 kJ/kg.