Answer to Question #258214 in Mechanics | Relativity for giel

Question

378 𝑙𝑝𝑠 of air is isothermally compressed at 27℃ and 1379 𝑘𝑃𝑎𝑎; 𝑝2 = 4136 𝑘𝑃𝑎𝑎. For both nonflow and steady-flow (𝑉1 = 23 𝑚 𝑠, 𝑉2 = 46 𝑚 𝑠, ∆𝑃 = 0) processes, compute (a) ∫ 𝑝𝑑𝑉 𝑎𝑛𝑑 − ∫ 𝑉𝑑𝑝 (b) ∆𝑈, ∆𝐻, 𝑎𝑛𝑑 ∆𝑆 (c) 𝑊 𝑎𝑛𝑑 𝑄

Answer 1

a)

$\int pdV =1379\cdot 10^3\displaystyle{\int^{46}_{23}}dV=1379\cdot 10^3\cdot 23=31717\cdot 10^3$ J

$\int Vdp =378\displaystyle{\int^{4136}_{1379}}dp=378\cdot 2757=1042146$ J

b)

$\Delta U=Q-W=0$

for non-flow:

$\Delta H=V\Delta p=1042146$ J

for steady-flow:

$\Delta H=p\Delta V=31717\cdot 10^3$ J

$\Delta S=Q/T$

$T=27-273=-246$ K

for non-flow:

$Q=\int pdV=31717\cdot 10^3$ J

$\Delta S=-31717\cdot 10^3/246=-129$ J

for steady-flow:

$Q=\int Vdp =1042146$ J

$\Delta S=-1042146/246=-4236$ J

c)

for non-flow:

$Q=W=\int pdV=31717\cdot 10^3$ J

for steady-flow:

$Q=W=\int Vdp=1042146$ J