A closed vessel contains gas at a pressure of 500 kPa. What pressure will be established in the vessel

A closed vessel contains gas at a pressure of 500 kPa. What pressure will be established in the vessel if, after opening the tap, 0.8 mass of gas leaves the vessels at a constant temperature?

Given: p1 = 500 kPa; dm = 0.8m1.

Find: p2.

Decision:

Let us write the Mendeleev-Clapeyron equation for two cases – before and after the gas is released from the vessel:

p1 * V = (m1 / M) * R * T;

p2 * V = (m2 / M) * R * T.

We divide the second equation by the first and express the required pressure p2:

p2 / p1 = m2 / m1;

p2 = p1 * (m2 / m1);

p2 = p1 * ((m1 – dm) / m1);

p2 = p1 * (1 – dm / m1);

p2 = 500 * (1 – 0.8) = 100 kPa.

Answer: p2 = 100 kPa.