# With isothermal compression of a gas with a mass of 2 kg, which is at 27 degrees under a pressure of 500 kPa

**With isothermal compression of a gas with a mass of 2 kg, which is at 27 degrees under a pressure of 500 kPa, the gas pressure increases 3 times. The compression work is 1.4 MJ. What gas was subjected to isothermal compression and what is its initial volume?**

Given:

m = 2kg;

t = 27 ° C;

p1 = 500kPa = 5 * 10 ^ 5 Pa;

p2 = 3 * p1;

T = const;

A = 1.4MJ = 1.4 * 10 ^ 6 J;

R = 8.31J / (mol * K) – universal gas constant;

Find: V1 and determine the type of gas.

In an isothermal process, the work of gas compression is determined from the formula:

A = p1 * V1 * ln (p2 / p1);

V1 = A / [p1 * ln (p2 / p1)] = (1.4 * 10 ^ 6) / [5 * 10 ^ 5 * ln (3)] = 2.55m ^ 3;

Let’s use the equation of state for an ideal gas:

p * V = (m / μ) * R * T,

Where

T = t + 273 ° = 300 ° K – absolute gas temperature,

μ is the molar weight of the gas;

From here we find the molar weight of the gas:

μ = m * R * T / (p1 * V1) = 2 * 8.31 * 300 / (5 * 10 ^ 5 * 2.55) = 0.004kg / mol = 4g / mol.

This gas is helium.