Question

A steel cylinder contains \(150.0\) moles of argon gas at a temperature of \(25^{\circ} \mathrm{C}\) and a pressure of \(8.93 \mathrm{MPa}\). After some argon has been used, the pressure is \(2.00 \mathrm{MPa}\) at a temperature of \(19^{\circ} \mathrm{C}\). What mass of argon remains in the cylinder?

Short answer

The mass of argon that remains in the cylinder after some of it has been used is approximately \(1342.42\, \mathrm{g}\).

Step-by-step solution

Write down the given information

We are given the initial conditions of the steel cylinder containing argon gas: - Moles of argon (n₁) = 150.0 moles - Temperature (T₁) = 25°C = 298.15 K (converted to Kelvin) - Pressure (P₁) = 8.93 MPa = 8.93 * 10^6 Pa (converted to Pascal) And the final conditions of the cylinder: - Temperature (T₂) = 19°C = 292.15 K (converted to Kelvin) - Pressure (P₂) = 2.00 MPa = 2.00 * 10^6 Pa (converted to Pascal) We are asked to find the mass of argon that remains in the cylinder.

Use the Ideal Gas Law equation

Recall that the Ideal Gas Law equation is: \( PV = nRT \) Where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant (8.314 J/mol K), and T is the temperature in Kelvin. First, let's find the volume of the initial state using the Ideal Gas Law: \( V₁ = \frac{n₁RT₁}{P₁} \) Substituting the given values, we have: \( V₁ = \frac{(150.0\, \mathrm{moles})(8.314\, \mathrm{J/mol\, K})(298.15\, \mathrm{K})}{8.93 \times 10^6\, \mathrm{Pa}} \)

Calculate volume in the initial state (V₁)

Evaluate the expression for V₁: \( V₁ = \frac{(150.0\, \mathrm{moles})(8.314\, \mathrm{J/mol\, K})(298.15\, \mathrm{K})}{8.93 \times 10^6\, \mathrm{Pa}} \approx 0.498\, \mathrm{m^3} \) So, the volume of the cylinder in the initial state is approximately 0.498 m³.

Calculate the number of moles of argon in the final state (n₂)

Since the volume of the cylinder remains constant, we can use the Ideal Gas Law equation for the final state and solve for n₂: \( P₂V = n₂RT₂ \Rightarrow n₂ = \frac{P₂V}{RT₂} \) Substituting the given values, we have: \( n₂ = \frac{(2.00 \times 10^6\, \mathrm{Pa})(0.498\, \mathrm{m^3})}{(8.314\, \mathrm{J/mol\, K})(292.15\, \mathrm{K})} \)

Calculate the value of n₂

Evaluate the expression for n₂: \( n₂ = \frac{(2.00 \times 10^6\, \mathrm{Pa})(0.498\, \mathrm{m^3})}{(8.314\, \mathrm{J/mol\, K})(292.15\, \mathrm{K})} \approx 33.62\, \mathrm{moles} \) So, there are approximately 33.62 moles of argon in the final state.

Convert moles to mass

To calculate the mass of argon in the final state, we need to multiply the number of moles by the molar mass of argon (39.95 g/mol): \( m_\mathrm{Ar} = n₂ \times M_\mathrm{Ar} = 33.62\, \mathrm{moles} \times 39.95\, \mathrm{g/mol} \) Evaluate the expression for m_Ar: \( m_\mathrm{Ar} = 33.62\, \mathrm{moles} \times 39.95\, \mathrm{g/mol} \approx 1342.42\, \mathrm{g} \) So, there are approximately 1342.42 g of argon remaining in the cylinder after some of it has been used.

Key concepts

Argon Gas Properties

Argon is a noble gas, which is a group of elements known for their lack of chemical reactivity. This inertness is due to the full valence shell of electrons that argon atoms possess, making them stable and unlikely to form chemical bonds. With an atomic number of 18, argon is both colorless and odorless in its natural state, and it does not readily combine with other elements or compounds.

In the context of our textbook exercise, the argon gas is contained within a steel cylinder. The physical properties of argon are important for understanding its behavior under various pressures and temperatures, as governed by the ideal gas law. Argon is mainly used in applications that require a non-reactive atmosphere, such as in incandescent light bulbs to protect the filament, in welding to prevent oxidation, and in double glazing windows for insulation. Under standard conditions, argon remains a gas, which is why it is usually stored under high pressure in steel cylinders.

Gas Law Calculations

One of the fundamental tools in understanding gas behavior is the Ideal Gas Law, represented by the equation \(PV = nRT\). This equation highlights the relationship between the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas, with R being the ideal gas constant.

In the exercise, we used the Ideal Gas Law to calculate the volume of argon gas in the cylinder at the initial conditions and the number of moles remaining after some argon was used. By knowing the initial amount of argon in moles and the conditions under which the pressure changed within a constant volume, we could deduce the final number of moles in the cylinder. Gas law calculations are essential in various scientific fields, including chemistry, physics, and engineering, because they help us predict and understand the behavior of gases under different environmental conditions.

Converting Moles to Mass

Understanding the relationship between moles and mass is crucial for many chemistry applications, including calculating the amount of a substance involved in a reaction or present in a container. Moles represent a quantity (Avogadro's number, 6.022 x 10^23 particles) of a substance, while mass is the measure of how much matter is present.

In our exercise, after determining the number of moles of argon remaining in the cylinder, we converted this number to mass using argon's molar mass. The molar mass of argon is 39.95 g/mol, which means each mole of argon has a mass of 39.95 grams. Multiplying the number of moles by the molar mass gives us the total mass of argon remaining in the cylinder. This conversion is essential for practical applications, where the actual mass of a substance is often more useful than the number of moles.