The cylinder represented above contains 2.2 kg of water vapor initially at a vol

Question

The cylinder represented above contains 2.2 kg of water vapor initially at a volume of 2.0 m3 and an absolute pressure of 3.0 x 105 Pa. This state is represented by point A in the PV diagram below. The molar mass of water is 18 g, and the water vapor can be treated as an ideal gas. (a) Calculate the temperature of the water vapor at point A. The absolute pressure of the water vapor is increased at constant volume to 4.0 x 105 Pa at point B, and then the volume of the water vapor is increased at constant pressure to 2.5 m3 at point C, as shown in the PV diagram. (b) Calculate the temperature of the water vapor at point C. (c) Does the internal energy of the water vapor for the process A?B?C increase, decrease, or remain the same? Justify your answer. (d) Calculate the work done on the water vapor for the process A?B?C.

Explanation / Answer

A)

First you must find the amount of moles contained in the gas.

2.2 kg of water vapor. 1 mol = 18 g. Therefore moles = (2.2*10^3)/18 = 122.222 moles.

Next finding the temperature. PV=nRT >>> T=PV/nR >>> T=(3*10^(5)*2)/(122.222*8.31)=590.744K

B) This can be done by PV=nRT again or PV/T=PV/T. For simplicity I'll do it with PV=nRT again.

T=PV/nR=(4*10^(5)*2.5)/(122.222*8.31)=984.574K

C) Increase. This can be justified by dQ=3/2nRdT >>> (d meaning delta)

dQ=3/2(122.222)(8.31)(984.574-590.744) = 600000J

Because the answer is positive, dQ increases.

D) W=-PdV You have to account for both changes so

W=-3.5*10^(5)(2-2) + -4.0*10^5(2.5-2)=-200000J

The reason I used 3.5*10^5 Pa is because that is the average pressure from A>>>B.

You could also do the area of the triangle.

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