Question

A piston-cylinder assembly contains air, initially at 2 bar, 300 K, and a volume of 2 m^3. The air undergoes a process to a state where the pressure is 1 bar, during which the pressure-volume relationship is pV = constant. Assuming ideal gas behavior for the air, determine the mass of the air, in kg, and the work and heat transfer, each in kJ.

Answer 1

The mass of the air is 4.645 kg.

The work done or heat transfer is 277.25 kj.

Step-by-step explanation:

In isothermal process PV=constant. Take air as an ideal gas. For air gas constant is 287 j/kgK.

Given:

Initial pressure of the gas is 2 bar.

Initial volume of the gas is 2 m³.

Initial temperature is 300 K.

Final pressure is 1 bar.

Calculation:

Step1

Apply ideal gas equation for air as follows:

PV=mRT

`2*10^(5)*2=m* 287*300`

m = 4.645 kg.

Thus, the mass of the air is 4.645 kg.

Step2

For isothermal process work done is same as heat transfer.

Work done or the heat transfer is calculated as follows:

`W=P_(i)V_(i)ln(P_(i))/(P_(f))`

`W=2*10^(5)*2* ln(2)/(1)`

`W=2.7725*10^(5)` j

Or,

W=277.25 kj.

Thus, the work done or heat transfer is 277.25 kj.