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

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.