A 0.490-mol sample of an ideal diatomic gas at 432 kPa and 330 K expands quasi-s

Question

A 0.490-mol sample of an ideal diatomic gas at 432 kPa and 330 K expands quasi-statically until the pressure decreases to 154 kPa. Find the final temperature and volume of the gas, the work done by the gas, and the heat absorbed by the gas if the expansion is the following. (a) isothermal final temperature Correct: 330 Your answer is correct. K volume of the gas Correct: 8.7297 Your answer is correct. L work done by the gas Correct: 1386.67 Your answer is correct. J heat absorbed Correct: 1386.67 Your answer is correct. J (b) adiabatic final temperature K volume of the gas L work done by the gas J heat absorbed Correct: 0 Your answer is correct. J As you can see, i have got 5 of the 8 answers, it is on part b that i need the most help, i would highly appreciate it if this could be answered.

Explanation / Answer

V1 = nRT / P1 = 0.49 * 8.314 * 330 / 432000 = 0.0031119 m^3 Now,for adiabatic process : P1*V1^1.4 = P2*V2^1.4 So, V2 = 0.0065014 m^3 = 6.5014 l T2 = P2 * V2 / nR = 245.76 K W = K[V2^(1-y) - V1^(1-y)] / (1-y) = 857.87 J Q = 0...in adiabatic process

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