Question

Two thermally insulated vessels are connected by a narrow tube fitted with a valve that is initially closed as shown in the figure below. One vessel of volume V1 = 15.6 L, contains oxygen at a temperature of T1 = 295 K and a pressure of P1 = 1.76 atm. The other vessel of volume V2 = 23.0 L contains oxygen at a temperature of T2 = 435 K and a pressure of P2 = 2.10 atm. When the valve is opened, the gases in the two vessels mix and the temperature and pressure become uniform throughout. Pistons locked in place Valve P P. 12 T2 T (a) What is the final temperature? K (b) What is the final pressure? atm

Answer 1

P = 176 atm P2 = 2.20 atm V - 15.6 L V₂ = 23L T = 295k T₂ = 45K n2 = 0o. Initially let ni and na of moles of gas in two vessels. using ideal gas equation Pr=ART and "2= P2V2 n = Divi RT RT2 after value is opened then let ni and na are the no. moles of gases in both respectively and p e final pressure and I is final temperature n'=pvi and na = PV2 As total no. of moles of gase must be cons. constant. therefore nitne = nit na T T

+ PV2 KT + A (litve) PIVI + PzV2 PV & Ti Дт. RT PV| P2 V2 Ti Tz POVI PzV2 Ті + ("+") alt - Tz 1.76X 15.6 2.10 x 23 + 295 435 (15.6) +(23) ato K P = 0.026 T gases Assuming the both U - U1+U2 14Tf = n,&T, that have some Cv [U= Inteund energy]

PM + c P2v2) 75.156 K- u If = nititha Tz miTo the top + lamang ht (mitm) P2V2 PUITP 2 V2 RI2 If PUI T. + Tz) (-768156)+(2.10 425) IK (1.76x15.6) (2.10823) 0.204 435 Tf = 371-2K final temperature Pulting in equation is P= (371-240:026) atm P= 9.65 atm atm final pressure. + 295