Question

A chemist has a certain amount of gas under a pressure of \(33.3 \mathrm{~atm} ;\) it occupies \(30 \ell\) at \(273^{\circ} \mathrm{C}\), For his research, however, the gas must be at standard conditions. Under standard conditions what will the volume of the gas be?

Short answer

The volume of the gas under standard conditions will be approximately 50 liters.

Step-by-step solution

Recall the ideal gas law

The ideal gas law is given by the equation: \(PV = nRT\) Where P represents pressure, V represents volume, n represents the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin. Note: Since the number of moles of gas and the gas constant won't change in this problem, we can focus on the pressure, volume, and temperature variables.

Convert the given temperature to Kelvin

We are given the initial temperature of the gas in Celsius, but we need the temperature in Kelvin for the ideal gas law equation. To convert Celsius to Kelvin, we can use the following formula: \( T(K) = T(°C) + 273\) Substitute temperature value in the equation: \( T(K) = 273^\circ C + 273 = 546 K\)

Recall standard conditions

Standard conditions are defined as pressure at 1 atm and temperature at 273 K (0°C). We will use these values to find the final volume of the gas under standard conditions.

Use the ideal gas law to find the initial number of moles of the gas

We can rearrange the ideal gas law equation to find the number of moles of the gas, using the initial pressure, volume, and temperature: \( n = \frac{PV}{RT} \) Now, substitute the given values: \( P = 33.3 atm, V = 30 L, T = 546 K, R = 0.0821 \frac{L*atm}{mol*K} \) \( n = \frac{(33.3)(30)}{(0.0821)(546)} = 2.237 mol \) (approximately)

Calculate the final volume under standard conditions

Now, we know the number of moles of the gas, and we can use it to find the volume of the gas under standard conditions. Using the ideal gas law equation again, but this time with standard conditions for pressure and temperature: \( PV = nRT \) \( V = \frac{nRT}{P} \) Substitute the values: \( n = 2.237 mol, P = 1 atm, T = 273 K, R = 0.0821 \frac{L*atm}{mol*K} \) \( V = \frac{(2.237)(0.0821)(273)}{1} = 50 L \) (approximately) The volume of the gas under standard conditions will be approximately 50 liters.

Key concepts

Standard Conditions

In chemistry, standard conditions are often used as a reference point for experiments and calculations. Standard conditions for gases are defined as a pressure of 1 atmosphere (atm) and a temperature of 273 Kelvin (0 degrees Celsius). These conditions provide a baseline to compare the behaviors of gases under different environmental conditions.
These conditions are essential when using the Ideal Gas Law, as they allow scientists to predict how a gas will behave when changing from any set environment to another. This uniformity simplifies calculations, particularly when adjusting for real-world variations in experimental data.
Whether you're a student new to chemical equations or an experienced chemist, understanding and using standard conditions is crucial for accurate gas calculations.

Pressure and Volume

Pressure and volume are two fundamental properties of gases that are closely interrelated. According to Boyle's Law, when temperature is held constant, the volume of a gas is inversely proportional to its pressure (\[ P \times V = constant \]). This means if you increase the pressure, the volume decreases, and vice versa.
In this exercise, the chemist measures gas at a high pressure (33.3 atm) and needs to convert this condition to standard pressure (1 atm). This requires rearranging the gas law equation to solve for the new volume.
These calculations are essential for accurately understanding the gas's behavior at different pressures, which is critical for numerous scientific and industrial applications.

Temperature Conversion

Temperature plays a vital role in gas behavior and must be correctly converted for analyses involving the Ideal Gas Law. Temperature is typically measured in Celsius or Fahrenheit, but for these calculations, Kelvin is the standard unit.
To convert from Celsius to Kelvin, simply add 273 to the Celsius temperature. In our exercise, the initial temperature was 273°C, giving a Kelvin equivalent of 546 K after conversion.
Why is Kelvin used here? Kelvin is an absolute temperature scale that starts from absolute zero, making it ideal for scientific calculations to avoid negative values that aren't meaningful in the context of gas laws.

• Kelvin eliminates the complications of Celsius that arise at near-freezing temperatures.
• It ensures consistency across different equations and calculations.

Moles of Gas

The mole is a fundamental concept in chemistry that measures substance quantity. When dealing with gases, knowing the number of moles (\( n \)) is crucial for applying the Ideal Gas Law. As per Avogadro's principle, equal volumes of gases at the same temperature and pressure contain the same number of moles.
In our exercise, we start by calculating the moles using the initial conditions given. Using the rearranged Ideal Gas Law equation, \( n = \frac{PV}{RT} \), the moles of gas are found to be approximately 2.237.
This value is then used alongside standard conditions to find the resulting volume of the gas. Having accurate mole calculations ensures precise results, which is important for not only theoretical but also practical applications in experiments and industrial gas processes.

• Always remember to use consistent units (atm for pressure, L for volume, K for temperature).
• The Ideal Gas Law relates moles to physical properties, linking microscopic and macroscopic observations.