Answer to Question #111160 in Physics for please

Question

https://upload.cc/i1/2020/04/13/Ei4lqp.jpg

question in picture

Answer 1

Solution.

a) $PV = nRT \implies n = PV/RT;$

$n = (1.55\sdot 10^5 Pa\sdot5m^3)/(8.31 J/mol K\sdot350K) = 266.5mol;$

$T_c = PV/nR;$

$Tc = (1.85\sdot10^5 Pa\sdot20m^3)/(266.5mol\sdot8.31J/molK) = 167K;$

b)$W_Bc = 0; W_CA = (P_A + P_C)/2\sdot(V_C-V_A);$

$W_CA = (1.55\sdot10^5 + 1.85\sdot10^5)/2\sdot(20-5) =25.5\sdot10^5J = 2.55MJ;$

c) $Q_BC = W_BC + \Delta E_int; \implies \Delta E_int = Q_BC - W_BC;$

$\Delta E_int = 5088660J;$

$Q_CA = W_CA + \Delta E_int; \implies \Delta E_int = Q_CA - W_CA;$

$\Delta E_int =-6937500 - 2.55\sdot10^6 = 9.49\sdot10^6J;$

d) $W_AB = -\Delta E_int; W_AB = -5550000J = -5.5\sdot10^6J;$

e) $W_ABC = W_AB + W_BC + W_CA;$

$W_ABCA = -5.5\sdot10^6 + 0 + 2.55\sdot10^6 = -2.95MJ;$

$Q_ABCA = Q_AB + Q_BC + Q_CA;$

$Q_ABCA = 0 + 5088660 -6937500 = -1848840J;$

$\Delta E_int_ABCA =\Delta E_int_AB + \Delta E_int_BC + \Delta E_int_CA;$

$\Delta E_int_ABCA = -5550000 + 5088660 -6937500 = 3701160J ;$

Answer: a)$n = 266.5 mol; T_C = 167K;$

b) $W_BC = 0; W_CA = 2.55 MJ;$

c) $\Delta E_int_BC = 5.088MJ; \Delta E_int_CA = 9.49MJ;$

d) $W_AB = -5.5MJ;$

e)$W_ABCA = -2.95MJ;$

$Q_ABCA = -1.85MJ;$

$\Delta$E_int_ABCa = 3.7MJ.