What should be the smallest volume of the cylinder so that it can hold 5 kg

What should be the smallest volume of the cylinder so that it can hold 5 kg of oxygen at a temperature of 10 ° C if its walls withstand a pressure of 15 MPa?

m = 5 kg.

t = 10 ° C.

Pmax = 15 MPa = 15 * 10 ^ 6 Pa.

R = 8.31 m2 * kg / s2 * ° K * mol.

M (O2) = 0.032 kg / mol.

Vmin -?

The oxygen pressure in the cylinder P must not exceed the maximum pressure on the walls that they withstand Pmax: P <Pmax.

Let us find the volume of oxygen Vmin at Pmax from the Mendeleev-Cliperon equation: Pmax * Vmin = m * R * T / M.

Vmin = m * R * T / M * Pmax.

Since oxygen is a diatomic gas O2, its molar mass is M (O2) = 0.032 kg / mol.

We find the absolute temperature T by the formula: T = 273 + t = 273 + 10 = 283 ° K.

Vmin = 5 kg * 8.31 m2 * kg / s2 * ° K * mol * 283 ° K / 0.032 kg / mol * 15 * 10 ^ 6 Pa = 0.025 m3.

Answer: to prevent an explosion, the smallest volume of the cylinder should be Vmin = 0.025 m3.



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