Question

A 12.8 L cylinder contains \(35.8 \mathrm{g} \mathrm{O}_{2}\) at \(46^{\circ} \mathrm{C}\). What is the pressure of this gas, in atmospheres?

Short answer

The pressure of the gas is \(2.29 \, \mathrm{atm}\).

Step-by-step solution

Convert grams of \(O_2\) to moles

We first need to convert the mass of \(O_2\) (oxygen) in grams into moles, by using the molar mass of \(O_2\), which is \(32.00 \, \mathrm{g/mol}\). This is done using the formula \(n = m/M\), where \(m\) is the mass and \(M\) is the molar mass. This gives us: \[n = 35.8 \, \mathrm{g} / 32.00 \, \mathrm{g/mol} = 1.119 \, \mathrm{mol}\]

Convert temperature from Celsius to Kelvin

We need to convert the temperature from Celsius to Kelvin, because the ideal gas law uses the Kelvin temperature scale. We can do this using the formula \(T(K) = T(C) + 273.15\), which gives us: \[T = 46^{\circ}C + 273.15 = 319.15 \, K\]

Substitute values into the ideal gas law and solve for P

Next, we put these values into the ideal gas law equation \(PV = nRT\), and solve for the pressure \(P\). Remembering that the ideal gas constant \(R = 0.0821 \, \mathrm{(atm \cdot L)/(mol \cdot K)}\), we get: \[P = \frac{nRT}{V} = \frac{1.119 \, \mathrm{mol} \cdot 0.0821 \, \mathrm{(atm \cdot L)/(mol \cdot K)} \cdot 319.15 \, K}{12.8 \, L} = 2.29 \, \mathrm{atm}\] This pressure is the final answer to the exercise.

Key concepts

Molar Mass

In the context of chemistry, molar mass is a crucial concept. It represents the mass of one mole of a given substance. For oxygen (O_2), the molar mass is 32.00 \, \mathrm{g/mol}. This receives its value from the sum of the atomic masses of two oxygen atoms. Calculating moles from a given mass involves the formula \(n = \frac{m}{M}\), where:

• \(n\) is the number of moles.
• \(m\) is the mass in grams.
• \(M\) is the molar mass in \(\mathrm{g/mol}\).

This equation forms the basis for many calculations in chemistry. For example, if you have 35.8 grams of O_2, dividing this by 32.00 \, \mathrm{g/mol} yields approximately 1.119 moles. This conversion helps in elucidating other properties of gases, such as calculations involving the Ideal Gas Law.
Understanding the basis of molar mass allows you to bridge the gap between macroscopic measurements and molecular quantities. This is essential for experiments and theoretical calculations in chemistry.

Temperature Conversion

Temperature conversion is a fundamental task when dealing with gas laws. In science, temperatures must often be converted to Kelvin. This is due to the Kelvin scale being absolute, which aligns with the equations of thermodynamics, including the Ideal Gas Law.
To convert from Celsius to Kelvin, you use the formula:

• \(T(K) = T(C) + 273.15\).

This simple addition helps you align your calculations with scientific standards. For instance, a temperature of 46^{\circ}C converts to 319.15 \, K.
Utilizing the Kelvin scale ensures that all thermal energy values remain proportional, avoiding the pitfalls of scales that include negative values, such as Celsius. Maintaining consistency in temperature calculations ensures the accuracy of gas law equations, critical for achieving precise results in chemical experiments.

Gas Pressure Calculation

Gas pressure calculation is an important part of understanding how gases behave. The Ideal Gas Law is a comprehensive formula that relates the pressure (P), volume (V), and temperature (T) of a gas to the number of moles (n) and the gas constant (R).

• The formula is expressed as \(PV = nRT\).
• \(P\) is the pressure in atmospheres.
• \(V\) is the volume in liters.
• \(n\) is the number of moles.
• \(R\) is the ideal gas constant, 0.0821 \, \mathrm{(atm \cdot L)/(mol \cdot K)}.
• \(T\) is the temperature in Kelvin.

By rearranging this equation, you can solve for pressure: \(P = \frac{nRT}{V}\). With given values, such as 1.119 moles of O_2, a volume of 12.8 \, L, and a temperature of 319.15 \, K, you can determine the pressure as approximately 2.29 \, atm.
This showcases the interdependence of temperature, volume, and moles in determining the state of a gas. Each parameter within the Ideal Gas Law holds significance in predicting and analyzing gas behaviors under various conditions.