Question

A Flexible Balloon A flexible balloon contains 0.315 mol of an unknown polyatomic gas. Initially the balloon containing the gas has a volume of 7400 cm and a temperature of 29.0°C. The gas first expands isobarically until the volume doubles. Then it expands adiabatically until the temperature returns to its initial value. Assume that the gas may be treated as an ideal gas with Cp = 33.26 J/mol. K and y = 4/3. Part A What is the total heat supplied to the gas in the process? You did not open hints for this part. ANSWER: J Part B What is the total change in the internal energy AU of the gas? You did not open hints for this part. ANSWER: ΔU = J Part C What is the total work W done by the gas? You did not open hints for this part ANSWER: W= J Part D What is the final volume V? You did not open hints for this part. ANSWER: V = m

Answer 1

in=0.315 male Р A PB=P Isobaric B Adia- -batic VA= 7400cm3 7400 x 10 m Pc A A B so 3 > A=7.4 x 103 m² V=2VA= 2x7.4x103 m3 > 98=1428X10-3 m3 Temperature at A is TA=29°C = 1273 +29)K ITA=302K Initial temperature. Since final temperature (Tc) is same

is same as eneitial temperature (Tw). So, T=302K ARTB TB Process AB i sobaric So, PAVA - ARTA V = TA (0 PA=PB) PB VB - TB= ( 16 J x TA = 2 VA x 302K VA TB=604K Part A Co= 33.26J/meil k Tatal heat supplied to gas is (Q= QAB + Qec QAB = nO (To-TA) CAB is iso - > QAR = 0.315x 33.264(604-302) J 3164.0238 I HE4/3 baric) AB QBC =0 Because BSC is adiabatic .: Q = (3164.0238 +0]J >Q= 3164.0238 ] Part B. Total change in internal energy is AU=nu (Tc-TA)=0

X 14.8x10 m3 AuzoJ Because (Tc = T). Partc from 1st low of thermodynamics Q= WTAU We Q-AU= Q-0 = Q ->W=3164.0238.J Part D for adiabeetic process Tuis constant .: To ug Te va 여 r-1 TB Ir Vc = TB To , "B sve = TB To NB 604 Vc · 302 Ve= x 14,8x103 m3 → Vo = 8x14.8x10-3 m3 IN=0.1184 m²