Question

A gas sample containing 1.50 moles at \(25^{\circ} \mathrm{C}\) exerts a pressure of 400. torr. Some gas is added to the same container and the temperature is increased to 50\(\cdot^{\circ} \mathrm{C}\). If the pressure increases to 800. torr, how many moles of gas were added to the container? Assume a constant-volume container.

Short answer

Approximately 0.7537 moles of gas were added to the container.

Step-by-step solution

Convert temperatures to Kelvin

Since the given temperatures are in Celsius, we first need to convert them to Kelvin. So, add 273.15 to each given temperature: T1 = 25 + 273.15 = 298.15 K T2 = 50 + 273.15 = 323.15 K

Convert pressure to atm

The given pressures are in torr unit, we need to convert them to atm unit, using the fact that 1 atm = 760 torr: P1 = 400 torr * (1 atm / 760 torr) = 0.5263 atm P2 = 800 torr * (1 atm / 760 torr) = 1.0526 atm

Use the Ideal Gas Law

Since the volume and gas constant R are constant, we can work with the ratios of pressure, moles, and temperature. \[\frac{P_1V}{n_1RT_1}=\frac{P_2V}{n_2RT_2}\] \[\frac{P_1}{n_1T_1} = \frac{P_2}{n_2T_2}\]

Plug in the known values

We can now plug in the known values for P1, n1, T1, P2, and T2, and solve for n2 (the moles after adding gas). \[\frac{0.5263}{1.50 \times 298.15} = \frac{1.0526}{n_2 \times 323.15}\]

Solve for n2

Solving for n2, we get: \[n_2 = \frac{1.0526 \times 1.50 \times 298.15 \times 323.15}{0.5263 \times 298.15}\] \[n_2 = 2.2537\]

Find the difference in moles

Now that we have n2 (the number of moles after adding gas), we can find the difference in the moles, which represents the moles of gas added to the container. moles_added = n2 - n1 moles_added = 2.2537 - 1.50 moles_added = 0.7537 moles So, approximately 0.7537 moles of gas were added to the container.

Key concepts

Temperature Conversion to Kelvin

Temperature conversion is a critical first step in many chemistry problems, especially when we're dealing with temperatures provided in degrees Celsius. The Kelvin scale is the SI unit for temperature and is commonly used in scientific calculations when applying the Ideal Gas Law. It's named after the physicist Lord Kelvin and starts from absolute zero, the theoretically lowest possible temperature.
To convert a temperature from Celsius to Kelvin, you simply add 273.15. This is because 0 degrees Celsius is equivalent to 273.15 Kelvin. For example, if you have a temperature of 25°C, the conversion is:

• Add 273.15 to the Celsius temperature:
• 25 + 273.15 = 298.15 K

This conversion is essential because gas laws, like the Ideal Gas Law, require temperature to be in Kelvin. When temperatures are in Kelvin, there's no need to worry about negative values which can complicate calculations.

Pressure Conversion to atm

In many chemistry problems, pressure needs to be converted from the given unit to atmospheres (atm), as this is a standard unit used to make calculations easier and consistent with gas law applications. The conversion factor between torr and atm is essential: 1 atm = 760 torr.
This conversion can be performed using the following relation:

• Divide the pressure in torr by 760:
• Suppose you have a pressure of 400 torr:
• 400 torr * (1 atm / 760 torr) = 0.5263 atm

For higher pressures like 800 torr, the process would be similar:

• 800 torr * (1 atm / 760 torr) = 1.0526 atm

By converting pressures to atm, it simplifies your calculations, especially when using the Ideal Gas Law equations and makes your results consistent with standard scientific notation.

Constant-Volume Process

In thermodynamics, a constant-volume process is where the volume of a system does not change. This is significant when applying the Ideal Gas Law (PV = nRT), as it implies that the volume, V, is a constant in calculations, simplifying the formula to focus more on the relations between pressure (P), number of moles (n), and temperature (T).
For a constant-volume process in an ideal gas scenario, if other variables like volume and the gas constant R don't change, the formula can be simplified to the ratio:

• \( \frac{P_1}{n_1T_1} = \frac{P_2}{n_2T_2} \)

This equation indicates that the ratio of pressure to the product of moles and temperature remains constant. It helps to solve problems like determining how many moles of gas have been added when pressure and temperature change but the container's size stays the same. This is powerful as it allows us to isolate variables and solve for unknowns under the constraint of constant volume.