Respuesta :
Explanation:
(a) The given data is as follows.
mass = 1 kg = 1000 g (as 1 kg = 1000 g)
Molar mass of [tex]NH_{3}[/tex] = 17 g/mol
[tex]P_{1}[/tex] = 2.5 bar = [tex]2.5 \times 10^{5} Pa[/tex] (as 1 bar = [tex]10^{5} Pa[/tex])
[tex]P_{2}[/tex] = 5 bar = [tex]5 \times 10^{5} Pa[/tex]
[tex]T_{1} = 30^{o}C[/tex] = 30 + 273 = 303 K
For adiabatic process, [tex]PV^{\gamma}[/tex] = constant = k
[tex]\gamma[/tex] = 1.33 = [tex]\frac{C_{p}}{C_{v}}[/tex]
[tex]P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}[/tex]
[tex](\frac{V_{2}}{V_{1}})^{\gamma} = \frac{P_{1}}{P_{2}}[/tex]
[tex]\frac{V_{2}}{V_{1}} = (\frac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}[/tex]
[tex]V_{2} = (\frac{2.5 \times 10^{5}}{5 \times 10^{5}})^{\frac{1}{1.33}} \times \frac{nRT_{1}}{P_{1}}[/tex] (as PV = nRT)
= [tex](\frac{2.5}{5})^{\frac{1}{1.33}} \times \frac{58.82 \times 8.314 J/mol K \times 303 K}{2.5 \times 10^{5}}[/tex]
= 0.352 [tex]m^{3}[/tex]
Also, w = [tex]\frac{P_{1}V_{1} - P_{2}V_{2}}{\gamma - 1}[/tex]
= [tex]\frac{2.5 \times 10^{5} \times 0.5927 - 5 \times 10^{5} \times 0.352}{1.33 - 1}[/tex]
= -84318.2 J
As 1 kJ = 1000 J. So, -84318.2 J = -84.318 kJ
Hence, the work required in kJ is -84318.2 J.
(b) It is known that for adiabatic system Q = 0,
[tex]\Delta U = Q - w[/tex]
[tex]nC_{v}dT[/tex] = -w
dT = [tex]\frac{-w}{nC_{v}}[/tex]
= [tex]\frac{84318.2}{58.82 \times 1.6}[/tex]
= 895.93 K
We known that dT = [tex]T_{1} - T_{2}[/tex]
so, 895.93 = 303 K - [tex]T_{2}[/tex]
[tex]T_{2}[/tex] = (895.93 - 303)K
= 592.93 K
= (592.93 - 273.15)^{o}C
= [tex]319.78^{o}C[/tex]
Hence, the final temperature is [tex]319.78^{o}C[/tex].
