Question

An ideal gas at \(7^{\circ} \mathrm{C}\) is in a spherical flexible container having a radius of 1.00 \(\mathrm{cm} .\) The gas is heated at constant pressure to \(88^{\circ} \mathrm{C}\) . Determine the radius of the spherical container after the gas is heated. [Volume of a sphere \(=(4 / 3) \pi r^{3} \cdot ]\)

Short answer

The radius of the spherical container increases from \(1.00 \, \mathrm{cm}\) to approximately \(1.126 \, \mathrm{cm}\) after heating the ideal gas from \(7^{\circ} \mathrm{C}\) to \(88^{\circ} \mathrm{C}\) at constant pressure.

Step-by-step solution

Write down the given variables

We are given the following information: - Initial temperature \(T_1\): \(7^{\circ} \mathrm{C}\) - Volume of the sphere \(V_1= (4 / 3) \pi r^3 = (4/3) \pi (1.00 \, \mathrm{cm})^3\), where \(r_1 = 1.00 \, \mathrm{cm}\) - Final temperature, \(T_2 = 88^{\circ} \mathrm{C}\) We also know that the gas is heated at constant pressure.

Convert temperatures to Kelvin

To work with temperatures in an ideal gas law problem, we need to convert the Celsius temperatures to Kelvin. The conversion formula is: \(T(K) = T(^\circ C) + 273.15\) For the initial temperature, \(T_1(K) = 7 + 273.15 = 280.15 \, \mathrm{K}\) For the final temperature, \(T_2(K) = 88 + 273.15 = 361.15 \, \mathrm{K}\)

Use the relationship between initial and final volumes and temperatures

Since we are dealing with an ideal gas and the process happens under constant pressure, we can use the relationship: \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\] We know that: \(V_1 = (4 / 3) \pi (1.00 \, \mathrm{cm})^3\) \(T_1 = 280.15 \, \mathrm{K}\) \(T_2 = 361.15 \, \mathrm{K}\) We want to find \(V_2\), which is the volume of the gas after heating. Rearranging the relationship equation to solve for \(V_2\), we get: \[V_2 = \frac{V_1 \times T_2}{T_1}\]

Calculate the final volume \(V_2\)

Substitute the given values and calculate \(V_2\): \[V_2 = \frac{((4 / 3) \pi (1.00 \, \mathrm{cm})^3) \times 361.15 \, \mathrm{K}}{280.15 \, \mathrm{K}}\] After evaluating this expression, we get: \(V_2 \approx 4.154 \, \mathrm{cm^3}\)

Calculate the final radius \(r_2\)

Now that we have the final volume \(V_2\), we can use the volume formula for a sphere to find the final radius \(r_2\): \[V_2 = \frac{4}{3} \pi r_2^3\] Rearranging the equation to solve for \(r_2\), we get: \[r_2 = \sqrt[3]{\frac{3V_2}{4\pi}}\] Now, substitute the value of \(V_2\): \[r_2 = \sqrt[3]{\frac{3(4.154 \, \mathrm{cm^3})}{4\pi}}\] Evaluating this expression, we find that the final radius \(r_2\) is approximately: \(r_2 \approx 1.126 \, \mathrm{cm}\) Therefore, after heating the gas, the radius of the spherical container is approximately \(1.126 \, \mathrm{cm}\).

Key concepts

Volume of a Sphere

To find the volume of a sphere, we use the formula \[ V = \frac{4}{3} \pi r^3 \]Where:

• \(V\) is the volume of the sphere
• \(r\) is the radius of the sphere

This formula arises from integrating the surface of a sphere in 3D space. Calculating the volume involves cubing the radius, which emphasizes how even a small change in the radius can significantly affect the volume. This is applicable in our problem where the radius of the spherical container increases slightly due to heating, leading to a noticeable change in volume. It is important to always use consistent units when calculating, as any discrepancy can lead to incorrect results. For example, when the radius is in centimeters, the volume will be in cubic centimeters.

Temperature Conversion

Temperature conversion is crucial when dealing with gas laws. The Ideal Gas Law uses temperatures in Kelvin to maintain consistency and accuracy. The Kelvin scale starts at absolute zero, which is the lowest possible temperature where molecular motion nearly stops.
To convert Celsius to Kelvin, the formula is simple:\[ T(K) = T(^{\circ}C) + 273.15 \]
Using this conversion ensures the temperature is in an absolute form, enabling its direct use within calculations. In the exercise, both the initial and final temperatures are changed from Celsius to Kelvin:

• Initial: \(7^{\circ}C = 280.15 \, K\)
• Final: \(88^{\circ}C = 361.15 \, K\)

This conversion is crucial because the Ideal Gas Law and related equations require absolute temperatures to function correctly.
Always remember to double-check the conversions, as any error can significantly affect the results of the calculations.

Constant Pressure Process

A constant pressure process, known in thermodynamics as an isobaric process, involves the pressure of the gas remaining constant while other variables change. Under this process, the Ideal Gas Law relationship can be used: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where

• \(V_1\) and \(V_2\) are initial and final volumes respectively
• \(T_1\) and \(T_2\) are initial and final temperatures in Kelvin

This relationship translates to volume being directly proportional to temperature at constant pressure. As shown in the exercise, when the gas is heated from \(280.15 \, K\) to \(361.15 \, K\), the volume increases correspondingly. This results in the formula provided to find the final volume \(V_2\), which is pivotal in determining the final radius of the sphere after the gas heats up.
Understanding the behavior of gases under different thermodynamic processes, like constant pressure, is critical for accurately predicting and calculating outcomes in physics and chemistry.