Calculate the work done when 1.000 mole of a perfect gas expands reversibly from 1.0 L to 10 L at 298.0 K. Then, calculate the amount of work done when the gas expands irreversibly against a constant external pressure of 1.00 atm. Compare the two values and comment.

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Explanation:

For work done at constant temperature;

w = -nRTln(V2/V1)

V1 = 1.0 L

V2 = 10 L

w = -(1 * 8.314 * 298) ln(10/1)

w = -5704.8 J

When the gas expands irreversibly against a constant external pressure;

w = -(PΔV)

w = -(1.00 (10 - 1))

w = - 9 J

During expansion of a gas in which the initial and final limits of volume are given, the constant pressure process produces more work than the isothermal process as we have seen above.

answer: i dont know the answer but this might

explanation:

l

i think its 100.

hope this !