Question

Nitrogen at 100 psia and 300°F in a rigid container is cooled until its pressure is 10 psia. Determine the work done and the heat transferred during this process, in Btu/lbm. Take cv = 0.177 Btu/lbm·R at room temperature, for nitrogen.

Answer 1

Explanation:

Using the ideal gas equation, we will determine the final temperature as follows.

           $T_{2} = T_{1} \times \frac{P_{2}}{P_{1}}$

                       = $760 \times \frac{10}{100} R$

                       = 76 R

As the container is rigid, so no work is done. Therefore, heat loss is determined from the change in internal energies that will be obtained from table A-18E for the given temperatures.

                $q_{out} = -\Delta U$

                            = $-(u_{2} - u_{1})$

                            = $-\frac{1883.4 - 3774.9}{28.013}$ Btu/lbm

                            = 67.5 Btu/lbm

Therefore, we can conclude that work done is zero and heat transferred during this process, in Btu/lbm is 67.5 Btu/lbm.