Nitrogen at an initial state of 300 K, 150 kPa and 0.2 m3 is compressed slowly in an isothermal process to a final pressure of 800 kPa. Determine the work done to the nitrogen and the amount of heat transfer during this process.

Work = - 4175.23 J

Heat Transfer = -4175.23 J

Explanation:

The work done in an isothermal process, is given by the following formula:

W = RT ln (P1/P2)

where,

W = Work done

R = Universal Gas Constant = 8.314 J/mol.k

T = Constant Temperature = 300 K

P1 = initial pressure = 150 KPa

P2 = final pressure = 800 KPa

using these values, we get:

W = (8.314 J/mol.K)(300 k) ln (150/800)

W = - 4175.23 J

Here, negative sign shows that work is done on the system.

In isothermal process, from 1st law of thermodynamics:

Heat Transfer = Q = W     (Since change in internal energy is zero for isothermal processes)

W = - 4175.23 J

Here, negative sign shows that heat is transferred from the system to surrounding.

answer:

› 2-

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answer:

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