The gas in the vessel is under a pressure of 80kPa at a temperature of 300K.

The gas in the vessel is under a pressure of 80kPa at a temperature of 300K. How will the pressure change when the temperature drops to 100K?

Given:

p1 = 80 kPa.

T1 = 300 K.

T2 = 100 K.

Find: p2.

Let’s give the units of measurement in the SI system.

p1 = 80 kPa = 80,000 Pa.

Let’s write down the Mendeleev-Clapeyron equation.

For the starting temperature.

p1 * V = R * T1.

After the change.

p2 * V = R * T2.

Let’s write down what the volume of these formulas is.

V = (R * T1) / p1.

V = (R * T2) / p2.

Since both formulas are for the same volume of gas, we can equalize them.

(R * T1) / p1 = (R * T2) / p2.

Divide both sides of the formula by R.

T1 / p1 = T2 / p2.

Let’s find what the pressure p2 is equal to.

p2 = (p1 * T2) / T1.

p2 = (80000 * 100) / 300 = 26667 Pa = 26.667 kPa.

Answer: The pressure will drop to 26.667 kPa.