27.78| Grade Summary Submissions Problem 2: n = 7 moles of a monatomic ideal gas

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27.78|

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Problem 2: n = 7 moles of a monatomic ideal gas initially fills a V0 = 0.25 m3 container at P0 = 75 kPa. The gas undergoes an isobaric expansion to V1 = 1.1 m3. Next it undergoes an isovolumetric cooling to its initial temperature T0. Finally it undergoes an isothermal compressions to its initial pressure and volume.

Randomized Variablesn = 7 moles
V0 = 0.25 m3
P0 = 75 kPa
V1 = 1.1 m3 Part (a) Identify the P-V diagram that correctly represents this three step cycle. Part (b) Calculate the work done by the gas, W1, in kilojoules, during the isobaric expansion (first process). W1 =

27.78|

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Deductions 4% Potential 96%

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Attempts remaining: 9(4% per attempt) detailed view 1 4% Hints: 3% deduction per hint. Hints remaining: 3 Feedback: 5% deduction per feedback. Part (c) Calculate the heat absorbed Q1, in kilojoules, during the isobaric expansion (first process).
Part (d) Write an expression for the change in internal energy, ?U1 during the isobaric expansion (first process).
Part (e) Calculate the work done by the gas, W2, in kilojoules, during the isovolumetric cooling (second process).
Part (f) Calculate the heat absorbed Q2, in kilojoules, during the isovolumetric cooling (second process).
Part (g) Calculate the change in internal energy by the gas, ?U2, in kilojoules, during the isovolumetric cooling (second process).
Part (h) Calculate the work done by the gas, W3, in kilojoules, during the isothermal compression (third process).
Part (i) Calculate the change in internal energy, ?U3, in kilojoules, during the isothermal compression (third process).
Part (j) Calculate the heat absorbed Q3, in kilojoules, during the isothermal compressions (third process).

Explanation / Answer

b)

W12 = P0*(V2 - V1)

= 75*(1.1 - 0.25)

= 63.75 kJ

c)

Q = n*Cp*(T2 - T1)

But T2 - T1 = (P0*V2 - P0*V1) / (nR) = W / nR

Q = W*Cp / R

But Cp / R = gamma / (gamma - 1)

So, Q = W*gamma / (gamma - 1)

For monoatomic ideal gas, gamma = 1.67

Q = 63.75*1.67 / (1.67 - 1)

Q = 159.375 kJ

d)

E12 = Q - W

= 159.375 - 63.75

= 95.625 kJ

e)

W23 = 0 (since constant volume process)

f)

Initial temperature T0 = P0*V0 / (nR)

= 75*10^3 *0.25 / (7*R)

= 2678.57 / R

T2 = P2*V2 / (nR)

= 75*10^3 *1.1 / (7*R)

= 11785.7 / R

Q23 = n*(Cv)*(T3 - T2)

Q23 = 7*(3R/2)*(2678.57 / R - 11785.7 / R )

Q23 = -95625 J = -95.625 kJ

g)

U23 = Q23 - W23

= -95.625 kJ

h)

W3 = P3*V3 ln (V1 / V3)

= P1*V1 ln (V1 / V3)

= 75*0.25 ln (0.25 / 1.1)

= -27.78 kJ

i)

U3 = 0 (since isothermal process)

j)

Q3 = W3 = -27.78 kJ

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