A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is −0.3 kJ/K. Assuming the ideal gas model for the gas and negligible kinetic and potential energy effects, evaluate the work, in kJ.

W = -120 KJ

Explanation:

Since the piston–cylinder assembly undergoes an isothermal process, then the temperature is constant.

Thus; T1 = T2 = 400K

change in entropy; ΔS = −0.3 kJ/K

Formula for change in entropy is written as;

ΔS = Q/T

Where Q is amount of heat transferred.

Thus;

Q = ΔS × T

Q = -0.3 × 400

Q = -120 KJ

From the first law of thermodynamics, we can find the workdone from;

Q = ΔU + W

Where;

ΔU is Change in the internal energy

W = Work done

Now, since it's an ideal gas model, the change in internal energy is expressed as;

ΔU = m•C_v•ΔT

Where;

m is mass

C_v is heat capacity at constant volume

ΔT is change in temperature

Now, since it's an isothermal process where temperature is constant, then;

ΔT = T2 - T1 = 0

Thus;

ΔU = m•C_v•ΔT = 0

ΔU = 0

From earlier;

Q = ΔU + W

Thus;

-120 = 0+ W

W = -120 KJ

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