Question 6.185: A device brings 2 kg of ammonia from 150 kPa, -20°C to 400 k...

C.V. Ammonia of constant mass m_{2}=m_{1}=m out to source.

Energy Eq.3.5:                m \left( u _{2}- u _{1}\right)={ }_{1} Q _{2}-{ }_{1} W _{2}

Entropy Eq.6.37:            m \left( s _{2}- s _{1}\right)=\int d Q / T +{ }_{1} S _{2  gen }={ }_{1} Q _{2} / T +{ }_{1} S _{2  gen}

Process:    {P _{1} v _{1}}^{ n }={ P _{2} v _{2}}^{ n }        Eq. (8.27)

State 1: Table B.2.2

State 2: Table B.2.2

The work term is integration of PdV as done in text leading to Eq.6.29

Notice we did not use Pv = RT as we used the ammonia tables.

From Eq.6.37

Notice:
n = 1.54, k = 1.3
n > k

…………………………………

Eq.3.5: E_{2}-E_{1}={ }_{1} Q_{2}-{ }_{1} W_{2}

Eq.6.37 : S_{2}-S_{1}=\int_{1}^{2} d S=\int_{1}^{2} \frac{\delta Q}{T}+{ }_{1} S_{2 gen}

Eq.8.27 : \phi =(e−T_{0}s+ P_{0}v)−(e_{0} −T_{0}s_{0} + P_{0}v_{0})

Eq.6.29 :
\begin{aligned}{ }_{1} W_{2} &=\int_{1}^{2} P d V=\text { constant } \int_{1}^{2} \frac{d V}{V^{n}} \\&=\frac{P_{2} V_{2}-P_{1} V_{1}}{1-n}=\frac{m R\left(T_{2}-T_{1}\right)}{1-n}\end{aligned}