Consider the system below. Two tanks are connected by a valve that is initially
ID: 491368 • Letter: C
Question
Consider the system below. Two tanks are connected by a valve that is initially closed. At this closed state, the left tank contains carbon dioxide (CO2), at a mass of 2 kg, a temperature of 77 °C, and pressure of 0.7 atm. The right tank contains CO2 as well, but at a mass of 8 kg, a temperature of 27 °C, and a pressure of 1.2 atm. At some point, the value is opened so that the CO2 in the different tanks can mix. During mixing, the tanks may also receive heat from the surroundings. If the final mixed, equilibrium temperature of the CO2 is 42 °C, what is the (a) final pressure of the system, and (b) the heat transfer during the mixing process? (The specific heat of CO2 is 0.745 kJ/(kg-K).)
CO2 8 kg, 27 °C, 1.2 atm valve CO2 2 kg, 77 °C, 0.7 atmExplanation / Answer
Moles= mass/molar mass, molar mass of CO2= 44, moles of CO2 in tank-1= 8*1000/44 gmoles =181.82 , moles of CO2 in tank-2= 2*1000/44= 45.45
Gas law equation is PV=nRT, R= 0.0821 L.atm/mole.K
Volume before mixing : tank-1= nRT/P= 181.82*0.0821*(27+273) /1.2 =3732 L
Tank-2 = nRT/P= 45.45*0.0821*(77+273)/0.7=1866 L
Total volume after mixing = 3732+1866= 5598L
Final temperature, T2= 42 deg.c= 42+273= 315 K, moles after mixing = 181.2+45.45=226.65
From gas law equation, P= nRT/V= 226.25*0.0821*315/5598=1.04 atm
Entropy change in tank-1= m( mass of CO2 in tank-1)Cp*ln (T2/300) and entropy change in tank-2= m( mass of CO2 in tank-2)*Cp*ln(T2/350)
Hence entropy change in tank-1 = 8*0.745*ln(315/300)=0.290Kj/K
Entropy change in tank-2 = 2*0.745*ln(315/350)= -0.15699
Total entropy change =0.290-0.15699 = 0.133 Kj/K
Heat transferred = T2*deltaS= 0.133*315 KJ=41.2 KJ