Three moles of ideal gas is inside a sealed rectangular container of size 0.2 m x 0.2 m x 0.4 m. If the pressure inside the container is 498,840 N/m^2, what is the temperature of the gas inside the container?

320 K

Explanation:

Given:

Number of moles of gas, n = 3

Dimensions of the rectangular container = 0.2 m * 0.2 m * 0.4 m

Thus,

Volume of the container = 0.2 m * 0.2 m * 0.4 m = 0.016 m³

Pressure inside the container, P = 498,840 N/m²

Now,

From the ideal gas law, we have

PV = nRT

here,

T is the temperature

R is the ideal gas constant = 8.314 J / (mol·K)

on substituting the values, we get

498,840 * 0.016 = 3 * 8.314 * T

or

T = 320 K

Hence, the pressure inside the container is 320 K