Problem 15 - University Physics with Modern Physics

Question

A metal tank with volume 3.10 L will burst if the absolute pressure of the gas it contains exceeds 100 atm. (a) If 11.0 mol of an ideal gas is put into the tank at a temperature of \(23.0^{\circ} \mathrm{C},\) to what temperature can the gas be warmed before the tank ruptures? You can ignore the thermal expansion of the tank. (b) Based on your answer to part (a), is it reasonable to ignore the thermal expansion of the tank? Explain.

Step-by-Step Solution

Answer

The gas can be warmed to approximately 72.77°C before the tank ruptures. Ignoring thermal expansion is reasonable due to the small temperature difference.

Step 1: Understand the Known Values

First, identify what's given in the problem:- Volume of the tank, \( V = 3.10 \) L (convert to m³: \( 3.10 \times 10^{-3} \) m³)- Number of moles, \( n = 11.0 \) mol- Initial temperature, \( T_1 = 23.0^{\circ} \text{C} = 296.15 \) K (Convert Celsius to Kelvin)- Maximum pressure before rupture, \( P_2 = 100 \) atm (convert to Pascals: \( 100 \times 1.01325 \times 10^5 \) Pa) The Universal gas constant, \( R = 8.314 \) J/(mol K).

Step 2: Use the Ideal Gas Law Equation

Use the ideal gas law to state that at the initial condition, \( P_1V = nRT_1 \). We need to find \( P_1 \), the initial pressure:\[P_1V = nRT_1 \]Substitute the known values:\[P_1 = \frac{nRT_1}{V} = \frac{11.0 \times 8.314 \times 296.15}{3.10 \times 10^{-3}} \]

Step 3: Calculate Initial Pressure

Calculate \( P_1 \) using the values from Step 2:\[P_1 = \frac{11.0 \times 8.314 \times 296.15}{3.10 \times 10^{-3}} \approx 8.75 \times 10^6 \, \text{Pa} \]Now we have the initial pressure.

Step 4: Determine the Final Temperature Before Rupture

From the initial condition and ideal gas law, we use the final condition at rupture pressure \( P_2 \):\[P_2 = \frac{nRT_2}{V}\]Using \( T_2 \), the temperature at which the tank ruptures, solve for \( T_2 \):\[T_2 = \frac{P_2 \cdot V}{nR} = \frac{(100 \times 1.01325 \times 10^5) \cdot (3.10 \times 10^{-3})}{11.0 \times 8.314}\]

Step 5: Calculate Final Temperature

Perform the calculation for \( T_2 \):\[T_2 = \frac{100 \times 1.01325 \times 10^5 \times 3.10 \times 10^{-3}}{11.0 \times 8.314} \approx 345.92 \, \text{K} \]Convert back to Celsius:\[T_2 \approx 345.92 - 273.15 \approx 72.77^{\circ} \text{C}\]

Step 6: Evaluate Thermal Expansion Consideration

Based on \( T_2 \approx 72.77^{\circ}\text{C}\), evaluate the assumption:Since the difference between the initial and final temperature is relatively small (around 50°C), the volume change due to thermal expansion of the tank is minimal, thus reasonable to ignore in the calculation.

Key Concepts

Gas Pressure Calculations

When dealing with gases, pressure is a key factor to consider, particularly within a confined space like a tank. Pressure, in simple terms, is the force exerted by the gas particles when they collide with the walls of their container. For our problem, it's crucial to understand that the tank will burst if the pressure exceeds a certain limit, in this case, 100 atm.

The Ideal Gas Law, represented by the equation \( PV = nRT \), helps us understand the relationship between pressure \( P \), volume \( V \), and temperature \( T \), with \( n \) being the number of moles of the gas and \( R \) the universal gas constant. It's used to determine the conditions under which our gas operates without leading to a tank rupture. We calculate initial pressure \( P_1 \) using volume, moles, and initial temperature to ensure it doesn't exceed the rupture pressure during thermal adjustments.

In the solution, the Ideal Gas Law allowed us to calculate the initial pressure and see how much it changes with temperature. Remember to always keep an eye on the units; changing atm to Pascal or Celsius to Kelvin is a crucial step that can change results dramatically.

Thermal Expansion

Thermal expansion is a phenomenon where materials expand upon heating. For gases, this means increased motion and collisions at higher temperatures that can raise pressure in a confined tank. However, in our exercise, we are instructed to ignore the tank's thermal expansion.

Why disregard it? If you calculate the final temperature of the gas and compare it to the initial temperature, the difference is small enough (only about 50°C) to have a negligible impact on the tank itself in reality. Materials like steel, often used for such tanks, have a relatively low coefficient of expansion, indicating very little dimensional change over such a temperature range.

Therefore, in this context, we prioritize the gas properties and calculations, as the tank's expansion would not significantly alter the results of pressure at the bursting point. Understanding when and why certain aspects can be ignored helps simplify complex problems without losing accuracy in approximations.

Temperature Conversions

Temperature plays a crucial role in gas calculations. When using the Ideal Gas Law, all temperature values must be in Kelvin (K). This is because Kelvin starts at absolute zero, the theoretically lowest possible temperature where particles have minimal thermal motion, making calculations mathematically and physically consistent.

To convert Celsius to Kelvin, you use the simple formula:

• \( T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \)

This conversion was vital in our solution, as the gas initially was at 23.0°C, which is 296.15 K. Similarly, after increasing the temperature to when it bursts the tank at 345.92 K, converting back to Celsius helps understand the practical difference.

Keep in mind: always convert to Kelvin before inserting temperatures into any gas law equations, to ensure accuracy and avoid errors derived from incorrect scales.