Problem 10.79 Challenge

กำหนดให้ \(n = 2.00 mol, P_{1} = 100 kPa, T_{1} = 300 K\)

และ \(P_{2} = 120 kPa\)

a) จาก \(P_{1}V_{1} = nRT_{1}\)

\(V_{1} =\frac{nRT_{1}}{P_{1}}\)

\(V_{1} = 0.0499 m^{3} \;\blacksquare\)

\(_{\gamma\gamma}\)

b)จาก\(P_{1}V_{1}=P_{2}V_{2}\)

\(_{1/\gamma}\)

\(V_{2} = V_{1}(\frac{P_{1}}{P_{2}})\)

จาก \(c_{V}=\frac{d}{2}R=\frac{3}{2}R; d = 3\)

จาก \(c_{P}= c_{V}+R=\frac{5}{2}R\)

จาก \(\gamma =\frac{c_{P}}{c^{V}}=\frac{5}{3}\)

\(_{3/5}\)

จะได้ \(V_{2} = V_{1}(\frac{P_{1}}{P_{2}})\)

\(V_{2} = 0.0447 m^{3} \;\blacksquare\)

c) จาก \(P_{2}V_{2} = nRT_{2}\)

\(T_{2} =\frac{P_{2}V_{2}}{nR}\)

\(T_{2} = 323 K \;\blacksquare\)

d) จาก \(ΔU = nc_{V}\Delta T\)

\(\Delta U=n(\frac{3}{2}R)(T_{2}-T_{1}); c_{V}=\frac{3}{2}R\)

\(\Delta U = 573 J \;\blacksquare\)

e) จาก Adiabatic Process

\(\therefore Q = 0 J \;\blacksquare\)

f) จาก \(ΔU = Q + W\)

\(\Delta U = 0 + W; Adiabatic Process\)

\(W = \Delta U\)

\(W = 573 J \;\blacksquare\)