A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this process, it absorbs 500 J of heat from the surroundings. Required:
a. Calculate the energy change of the gas, DU.
b. Calculate the work, w, done on the gas in an irreversible adiabatic (q =0) process connecting the same initial and final.

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

a.

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

b.

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

W = 810.6 J

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answer

the arrows to the right of earth should point up.

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