A mass of 1.15 kg of air at 111 kPa and 27 C is contained in a gas-tight, frictionless piston cylinder device. The air is now compressed to a final pressure of 0.84 MPa. During this process, heat is transferred from the air such that the temperature inside the cylinder remains constant. Calculate the work input (in kJ) during this process.

Work input W = -200.39 KJ

Explanation:

From the question, we are given;

m = 1.15 kg

Constant temperature T1 = T2 = 27 + 273 = 300k

Since the temperature is constant, we can say that the process is isothermal

P1 = 111 KPa

P2 = 0.84 MPa = 0.84 * 1000 KPa = 840 KPa

Now what we want to calculate is W1-2

Mathematically, for isothermal process;

W1-2 = mRTlnP1/P2

where R can be obtained from table and it is equal to 0.287 KJ/kg.k

Hence;

W1-2 = (1.15)(0.287)(300)(ln 111/840)

W1-2 = 99.015 * -2.023871690525 = -200.39 KJ

Kindly note that the value of the work is negative because work is done on the system and not by the system

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation:

no. dot product   =0 when vecs at right angles