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Nitrogen at an initial state of 300 K, 150 kPa, and 0.2 m3is compressed slowly in an isothermal process to a final pressure of 800 kPa. Determine the work done during this process.

User Jubei
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1 Answer

3 votes

Answer:


W=-251096.465\ J negativesign denotes thatthe work is consumed by the system.

Step-by-step explanation:

Given:

Isothermal process.

initial temperature,
T_1=300\ K

initial pressure,
P_1=150kPa

initial volume,
V_1=0.2\ m^3

final pressure,
P_2=800\ kPa

The work done during an isothermal process is given by:


W=P_1.V_1* ln((P_1)/(P_2) )


W=150* 1000* \ln(150)/(800)


W=-251096.465\ J negativesign denotes thatthe work is consumed by the system.

User Asaf Mesika
by
7.9k points
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