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- az_lenderLv 72 months agoFavorite Answer
On a pV diagram, you'll have something like a right triangle with its right angle at upper right, but the "hypotenuse" along the lower left will bulge outward. The process sequence will begin at the bottom vertex and will go counter-clockwise.

In the heating at constant volume, no "work" is done, but heat must be added in the amount of

n*Cv*delta-T = (20.78 J/molK)*(n)(delta-T).

Here n = (2500 J)/(8.314 J/molK * 288.3K) = 1.043 mol,

and delta-T = 4*(288.3 K) = 1153 K, so

Q = 24.99 kJ, and this is also the quantity of the increase in internal energy.

In the cooling at constant pressure, the work done on the gas is p*delta-V = (500000 Pa)(0.020 m^3)

= 10000 J or 10 kJ.

Since the temperature returns to its original value, the change in internal energy must be -24.99 kJ.

Therefore the heat lost must be 34.99 kJ.

In the isothermal expansion, there is no change in internal energy, as there is no "delta-T," but the gas does do work on the environment, equivalent to the heat that must be added to cause the expansion. This is the integral of p dV, which is nRT ln(V2/V1)

= (p1)(V1) ln(V2/V1) = (2500 J)*ln(5) = 4.024 kJ.

(Sorry I keep making some error when I try to do the enthalpy.)

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