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Chapter 5: Problem 66

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

Expert verified
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

01

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.
02

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, \(T_2(K) = 88 + 273.15 = 361.15 \, \mathrm{K}\)
03

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: \[V_2 = \frac{V_1 \times T_2}{T_1}\]
04

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: \(V_2 \approx 4.154 \, \mathrm{cm^3}\)
05

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 \(1.126 \, \mathrm{cm}\).

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Most popular questions from this chapter

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