220k views
0 votes
A 2-kg mass of helium is maintained at 300 kPa and 278C in a rigid container. How large is the container, in m3?

1 Answer

5 votes

Final answer:

The volume of the container is approximately 7.82 × 10^-24 m^3.

Step-by-step explanation:

To find the volume of the container, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin. To do this, we add 273 to the given temperature: 278 + 273 = 551K. Next, we can rearrange the ideal gas law equation to solve for V:

V = (nRT) / P

In this case, n = molar mass / mass of helium atom = 4.0026 × 10^-3 kg/mol / 6.02 × 10^23 mol = 6.64 × 10^-27 kg.

Substituting the given values, we get:

V = (6.64 × 10^-27 kg × 8.314 J/(mol·K) × 551K) / (300,000 Pa)

V ≈ 7.82 × 10^-24 m^3

User Chintan S
by
7.6k points