Use the exact values you enter to make later calculations. A certain ideal gas h

Question

Use the exact values you enter to make later calculations. A certain ideal gas has a molar specific heat of Cv, = 7/2R. A 2.62-mol sample of the gas always starts at pressure 1.20 X 10^5 Pa and temperature 260 K. For each of the following processes, determine the final pressure (Pf), the final volume (Vf), the final temperature (Tf), the change in internal energy of the gas (triangle E int), the energy added to the gas by heat (Q), and the work done on the gas (W). (i) The gas is heated at constant pressure to 395 K. Your response differs from the correct answer by more than lO%. Double check your calculations. J (ii) The gas s heated at constant volume to 395 K. (iii) The gas is compressed at constant temperature to 150 kPa. The correct answer is not zero. J (iv) The gas is compressed adiabatically to 150 kPa.

Explanation / Answer

Note : I am solving the parts which are not done by you

Cv = 3.5 R

Cp = Cv + R = 4.5 R

no of moles = n = 2.62

P1 = 1.2 x 10^5 Pa

T1 = 260 K

(i) At COnstant Pressure , heated to 395 K

Q = n Cp Delta T = 2.62 x 4.5 x 8.314 x (395-260) = 13232.978 J

W = P Delta V = n R Delta T = 2.62 x 8.314 x (395-260) = 2940.661 J

Delta E = Q - W = 13232.978 - 2940.661 = 10292.317 J

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(ii) Consatnt volume , heated to 395 K

PV = n R T

P1V1 / T1 = P2 V2 / T2

P2 = (T2/T1) P1 = (395/260) x 1.2 x 10^5 = 182307.69 Pa = 182.307 kPa

Pf = 182.307 kPa

Vf = Vi = n R T / Pi  = 2.62 x 8.314 x 260 / (1.2 x 10^5) = 0.04719 m3 = 47.19 L

Q = Delta Eint   = n Cv Delta T = 2.62 x 3.5 x 8.314 x (395 - 260) = 10292.316 J

Delta Eint  = 10292.316 J

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(iii) Consatnt temperature , compressed to 150 kPa

Vf = n R T / P = 2.62 x 8.314 x 260 / (150x1000) = 0.03775 m3 = 37.75 L

Q = n R T ln (Vf / Vi ) = 2.62 x 8.314 x 260 x ln(Pi / Pf) = 2.62 x 8.314 x 260 x ln(1.2 x 10^5 / (150x10^3)) = -1263.77 J

W = Q = -1263.77 J

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(iv)

Adiabatic , compresses to 150 kPa

Q = 0

PVk = constant where k = Cp/Cv = 9/7

Vf = Vi x (Pi / Pf)1/k   = 47.19 x (1.2 x 10^5 / (150x1000))^(7/9) = 39.67 L

Tf  = Pf Vf / n R = 150 x 1000 x 0.03967 / (2.62 x 8.314) = 91.058 K

Q = 0

W = (P1 V1 - P2V2)/ ( k - 1) = (1.2 x 10^5 x 0.04719 - 150 x 1000 x 0.03967) / ( (9/7) - 1 ) = -1006.95 J

Delta Eint  = W = -1006.95 J

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