Question

(8%) Problem 12: A monatomic ideal gas initially fills a container of volume V = 0.15 m at an initial pressure of P= 380 kPa and temperature T = 375 K. The gas undergoes an isobaric expansion to V2 = 0.65 m and then an isovolumetric heating to P2 = 520 kPa. ► 25% Part (a) Calculate the number of moles, n, contained in this ideal gas. n = 0.00861 Grade Summary Deductions 6% Potential 94% HOME d E sin cos t an) cotan() asin() acos() atan acotan sinh cosh() | tanh) cotanh0 Degrees Radians ( 7 8 ^ 4 5 1 * 1 2 + - 0 VO BACKSPACE 9 6 3 Submissions Attempts remaining: 7 2% per attempt) detailed view 2% → END CLEAR DEL Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 1% deduction per feedback. A 25% Part (b) Calculate the temperature of the gas, in kelvins, after it undergoes the isobaric expansion. A 25% Part (c) Calculate the change in entropy of the gas, in kilojoules per kelvin, as the material undergoes the isobaric expansion. A 25% Part (d) Calculate the change in entropy of the gas, in kilojoules per kelvin, as the material undergoes the isochoric process.

Answer 1

a) use ideal gas equation, P*V = n*R*T

==> n = P*V/(R*T)

= 380*10^3*0.15/(8.314*375)

= 18.3 moles <<<<<<<----------Answer

b) In Isobaric expansion, V/T = constant

V1/T1 = V2/T2

T2 = T1*(V2/V1)

= 375(0.65/0.15)

= 1625 K <<<<<<<<<-----------------Answer

c) change in entropy in isobaric expansion,

delta_S = n*Cp*ln(T2/T1)

= n*(3/2)*Rln(T2/T1)

= 18.3*(5/2)*8.314*ln(1625/375) (since, for monatomic gas, Cp = (5/2)*R)

= 558 J/K

= 0.558 kJ/K <<<<<<------------Answer

d) In isochoric process, P/T = constant

P2/T2 = P3/T3

T3 = T2*(P3/P2)

= 1625*(520/380)

= 2224 K

change in entropy in isochoric expansion,

delta_S = n*Cv*ln(T3/T2)

= n*(3/2)*R*ln(T3/T2) (since, for monatomic gas, Cv = (3/2)*R)

= 18.3*(3/2)*8.314*ln(2224/1625)

= 71.6 J/K

= 0.0716 kJ/K <<<<<<------------Answer