# In a 3000 liter tank there is propane (C3H8), the amount of which is 140 mol, and the temperature is 300 K

**In a 3000 liter tank there is propane (C3H8), the amount of which is 140 mol, and the temperature is 300 K. What pressure does the gas exert on the walls of the vessel?**

Data: V – volume of the considered tank with propane (V = 3000 l, in SI system V = 3 m3); ν is the amount of propane substance (ν = 140 mol); T – abs. temperature of propane in the tank (T = 300 K).

Const: R – univers. gas constant (R = 8.31 J / (mol * K)).

To find out the propane pressure on the walls of the reservoir under consideration, we apply the Mendeleev-Clapeyron zn: Px * V = m * R * T / MC3H8 = R * T * ν, from where we express: Px = R * T * ν / V.

Let’s perform the calculation: Tx = Px = R * T * ν / V = 8.31 * 300 * 140/3 = 116.34 * 10 ^ 3 Pa = 116.34 kPa.

Answer: The propane pressure on the walls of the tank under consideration is 116.34 kPa.