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Ch.5 - Gases
Chapter 5, Problem 123

A steel container of volume 0.35 L can withstand pressures up to 88 atm before exploding. What mass of helium can be stored in this container at 299 K?

Verified step by step guidance
1
Identify the ideal gas law equation: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature.
Rearrange the ideal gas law to solve for \( n \): \( n = \frac{PV}{RT} \).
Substitute the given values into the equation: \( P = 88 \text{ atm} \), \( V = 0.35 \text{ L} \), \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \), and \( T = 299 \text{ K} \).
Calculate the number of moles of helium using the rearranged equation.
Convert the number of moles of helium to mass using the molar mass of helium (4.00 g/mol).

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin. This law allows us to calculate the amount of gas that can be contained in a given volume under specific conditions.
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Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). For helium, the molar mass is approximately 4.00 g/mol. Understanding molar mass is essential for converting between the mass of a gas and the number of moles, which is necessary for applying the Ideal Gas Law in practical calculations.
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Gas Pressure

Gas pressure is the force exerted by gas molecules colliding with the walls of their container. It is measured in atmospheres (atm) or other units such as pascals (Pa). In this scenario, the maximum pressure the container can withstand (88 atm) is crucial for determining the conditions under which helium can be stored safely without exceeding this limit.
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