A vessel with a capacity of 20 liters (in cubic liters), filled with air at a pressure of 0.4 MPa, is connected

A vessel with a capacity of 20 liters (in cubic liters), filled with air at a pressure of 0.4 MPa, is connected to a vessel from which the air has been removed. In this case, the pressure in both vessels becomes the same and equal to 1.0 * 10 ^ 5 Pa. Determine the capacity of the second vessel. The process is isothermal.

Given:

V1 = 20 liters = 0.02 m ^ 3 – the volume of the first vessel;

P1 = 0.4 MPa = 0.4 * 10 ^ 6 Pascals – initial air pressure;

P2 = 1 * 10 ^ 5 Pascals – pressure after unification of vessels.

It is required to determine the volume of the second vessel V2 (m ^ 3).

Since, according to the condition of the problem, the process is isothermal, then:

P1 * V1 = P2 * (V1 + V2), where V1 + V2 is the total volume after the vessels are combined.

P1 * V1 = P2 * V1 + P2 * V2;

P2 * V2 = V1 * (P1 – P2);

V2 = V1 * (P1 – P2) / P2 = 0.02 * (4 * 10 ^ 5 – 1 * 10 ^ 5) / 1 * 10 ^ 5 =

= 0.02 * 3 * 10 ^ 5/1 * 10 ^ 5 = 0.02 * 3 = 0.06 m ^ 3 (60 liters).

Answer: the volume of the second vessel is 0.06 m ^ 3 (60 liters).