Problem 32
Question
What is the pressure, in pascals, exerted by \(1242 \mathrm{g}\) CO(g) when confined at \(-25^{\circ} \mathrm{C}\) to a cylindrical tank \(25.0 \mathrm{cm}\) in diameter and \(1.75 \mathrm{m}\) high?
Step-by-Step Solution
Verified Answer
The pressure exerted by the CO(g) in the tank is 115065 Pa or 115 kPa.
1Step 1: Calculate the number of moles
The molar mass of CO(g) is \(28.01 g/mol\). Using this, the number of moles (\(n\)) can be calculated using the formula \(n = \frac{mass}{molar \, mass}\), so \(n = \frac{1242g}{28.01 g/mol} = 44.34 mol\).
2Step 2: Calculate the volume of the tank
The volume of a cylinder is given by \(V= \pi r^2h\), where \(r\) is the radius, and \(h\) is the height. As the diameter is given, the radius \(r = \frac{d}{2} = \frac{25 cm}{2}\). Convert the radius from cm to m as \(r = 0.125 m\) because the volume must be in cubic meters to match the units of the gas constant R. The volume becomes \(V= \pi (0.125 m)^2 \times 1.75 m = 0.086 m^3\).
3Step 3: Convert temperature to Kelvin
The temperature is given in Celsius but must be in Kelvin for the Ideal Gas Law, which can be done using the formula \(T(K) = T(°C) + 273.15\). So the temperature is \(-25 °C + 273.15 = 248.15 K\).
4Step 4: Use the Ideal Gas Law
The Ideal Gas Law is \(P = \frac{nRT}{V}\), where P is pressure, R is the Ideal gas constant (R = \(8.314 \, J/(mol \cdot K)\)), n is the number of moles, T is the temperature in Kelvin, and V is the volume in cubic meters. Substituting the calculated values in gives \(P = \frac{44.34 mol \times 8.314 J/(mol \cdot K) \times 248.15 K}{0.086 m^3}\)
5Step 5: Calculate the pressure
Calculate the pressure by performing the arithmetic to find \(P = 115065 Pa\) or \(115 kPa\). Therefore, the pressure exerted by the CO(g) in the tank is \(115 kPa\) or \(115065 Pa\).
Key Concepts
Moles CalculationCylinder Volume CalculationTemperature Conversion to Kelvin
Moles Calculation
Understanding moles and how to calculate them is key in chemistry, especially when using the Ideal Gas Law. Moles measure the amount of a substance. To find the number of moles from a given mass, you use the formula: \( n = \frac{\text{mass}}{\text{molar mass}} \).
For example, in the exercise, the molar mass of carbon monoxide (CO) is \(28.01 \, g/mol\). Given a mass of \(1242 \, g\), the number of moles \(n\) is calculated by dividing the mass by the molar mass: \( n = \frac{1242 \, g}{28.01 \, g/mol} = 44.34 \, mol\).
This step is crucial because the Ideal Gas Law (\( PV = nRT \)) requires the number of moles to calculate the pressure of a gas. Remember, the molar mass is specific to each element or compound, so always ensure you have the correct values.
For example, in the exercise, the molar mass of carbon monoxide (CO) is \(28.01 \, g/mol\). Given a mass of \(1242 \, g\), the number of moles \(n\) is calculated by dividing the mass by the molar mass: \( n = \frac{1242 \, g}{28.01 \, g/mol} = 44.34 \, mol\).
This step is crucial because the Ideal Gas Law (\( PV = nRT \)) requires the number of moles to calculate the pressure of a gas. Remember, the molar mass is specific to each element or compound, so always ensure you have the correct values.
Cylinder Volume Calculation
Calculating the volume of the container is essential for determining pressure using the Ideal Gas Law. For a cylindrical tank, volume \(V\) is calculated with the formula: \( V = \pi r^2 h \).
Consider the given exercise: the cylinder has a diameter of \(25.0 \, cm\), which means the radius \(r\) is half of that, or \(12.5 \, cm\). It’s important to convert centimeters to meters for consistent units, so \(r = 0.125 \, m\).
The height \(h\) is given as \(1.75 \, m\). Plug these into the formula to get the volume: \[ V = \pi (0.125 \, m)^2 \times 1.75 \, m = 0.086 \, m^3 \]
Calculating this volume accurately is vital because it directly affects the calculation of pressure in the Ideal Gas Law. This ensures that all units are consistent with the gas constant \(R\), which uses cubic meters.
Consider the given exercise: the cylinder has a diameter of \(25.0 \, cm\), which means the radius \(r\) is half of that, or \(12.5 \, cm\). It’s important to convert centimeters to meters for consistent units, so \(r = 0.125 \, m\).
The height \(h\) is given as \(1.75 \, m\). Plug these into the formula to get the volume: \[ V = \pi (0.125 \, m)^2 \times 1.75 \, m = 0.086 \, m^3 \]
Calculating this volume accurately is vital because it directly affects the calculation of pressure in the Ideal Gas Law. This ensures that all units are consistent with the gas constant \(R\), which uses cubic meters.
Temperature Conversion to Kelvin
In the realm of gas calculations, temperature should always be in Kelvin. The Ideal Gas Law, \( PV = nRT \), requires this to maintain consistency with the gas constant \(R\), which operates on the Kelvin scale.
The conversion from Celsius to Kelvin is straightforward: add \(273.15\) to the Celsius temperature. For instance, in the exercise, the temperature is \(-25^{\circ} C\). To convert, simply do \(-25^{\circ} C + 273.15 = 248.15 \, K\).
This step is often overlooked but is essential for accurate calculations within the Ideal Gas framework. Always ensure temperature is expressed in Kelvin to avoid errors in your results.
The conversion from Celsius to Kelvin is straightforward: add \(273.15\) to the Celsius temperature. For instance, in the exercise, the temperature is \(-25^{\circ} C\). To convert, simply do \(-25^{\circ} C + 273.15 = 248.15 \, K\).
This step is often overlooked but is essential for accurate calculations within the Ideal Gas framework. Always ensure temperature is expressed in Kelvin to avoid errors in your results.
Other exercises in this chapter
Problem 29
\(\mathrm{Kr}(\mathrm{g})\) in a 18.5 L cylinder exerts a pressure of \(11.2 \mathrm{atm}\) at \(28.2^{\circ} \mathrm{C} .\) How many grams of gas are present?
View solution Problem 30
A 72.8 L constant-volume cylinder containing \(7.41 \mathrm{g}\) He is heated until the pressure reaches 3.50 atm. What is the final temperature in degrees Cels
View solution Problem 33
What is the molar volume of an ideal gas at (a) \(25^{\circ} \mathrm{C}\) and 1.00 atm; \((b) 100^{\circ} \mathrm{C}\) and 748 Torr?
View solution Problem 34
At what temperature is the molar volume of an ideal gas equal to \(22.4 \mathrm{L},\) if the pressure of the gas is \(2.5 \mathrm{atm} ?\)
View solution