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Chapter 5: Problem 58

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 \(\left.=(4 / 3) \pi r^{3} .\right]\)

Short Answer

Expert verified
The radius of the spherical container after the gas is heated is approximately \(1.23 cm\).

Step by step solution

01

Write down the Ideal Gas Law equation

The Ideal Gas Law equation is given by \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.
02

Convert the temperatures to Kelvin

We need to convert the given temperatures from degrees Celsius to Kelvin. The formula for the conversion is \(T(K) = T(°C) + 273.15\). Initial temperature, \(T_1 = 7°C + 273.15 = 280.15 K\) Final temperature, \(T_2 = 88°C + 273.15 = 361.15 K\)
03

Calculate the initial volume

The volume of a sphere is given by the formula: \(V = \frac{4}{3} \pi r^3\). Using the initial radius, \(r_1 = 1.00 cm\), we can calculate the initial volume, \(V_1\). \(V_1 = \frac{4}{3} \pi (1.00)^3 = \frac{4}{3} \pi cm^3\)
04

Use the constant pressure property

Since the pressure remains constant during the heating process, we can write the Ideal Gas Law equation for both initial and final states as follows: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)
05

Solve for the final volume

Substituting the values of \(V_1, T_1\), and \(T_2\) into the above equation, we can solve for the final volume, \(V_2\): \(\frac{\frac{4}{3} \pi cm^3}{280.15 K} = \frac{V_2}{361.15 K}\) After rearranging the equation, we find that: \(V_2 = \frac{4}{3} \pi cm^3 \times \frac{361.15}{280.15} \approx 1.89 \pi cm^3\)
06

Find the new radius of the container

We can find the final radius \(r_2\) using the formula for the volume of a sphere. Setting \(V_2 = 1.89 \pi cm^3\), we can solve for \(r_2\): \(\frac{4}{3} \pi r_2^3 = 1.89 \pi cm^3\) Dividing both sides by \(\frac{4}{3} \pi\), we get: \(r_2^3 = 1.89 cm^3\) Take the cube root of both sides: \(r_2 \approx 1.23 cm\) The radius of the spherical container after the gas is heated is approximately 1.23 cm.

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Most popular questions from this chapter

The partial pressure of \(\mathrm{CH}_{4}(g)\) is \(0.175\) atm and that of \(\mathrm{O}_{2}(g)\) is \(0.250\) atm in a mixture of the two gases. a. What is the mole fraction of each gas in the mixture? b. If the mixture occupies a volume of \(10.5 \mathrm{~L}\) at \(65^{\circ} \mathrm{C}\), calculate the total number of moles of gas in the mixture. c. Calculate the number of grams of each gas in the mixture.

Natural gas is a mixture of hydrocarbons, primarily methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right) .\) A typical mixture might have \(\chi_{\text {methane }}=\) \(0.915\) and \(X_{\text {ethane }}=0.085 .\) What are the partial pressures of the two gases in a \(15.00\) - L container of natural gas at \(20 .^{\circ} \mathrm{C}\) and \(1.44\) atm? Assuming complete combustion of both gases in the natural gas sample, what is the total mass of water formed?

Consider the following chemical equation. $$ 2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ If \(25.0 \mathrm{~mL} \mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?

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You have two containers each with 1 mol of xenon gas at \(15^{\circ} \mathrm{C}\). Container A has a volume of \(3.0 \mathrm{~L}\), and container \(\mathrm{B}\) has a volume of \(1.0 \mathrm{~L}\). Explain how the following quantities compare between the two containers. a. the average kinetic energy of the \(\mathrm{Xe}\) atoms b. the force with which the Xe atoms collide with the container walls c. the root mean square velocity of the Xe atoms d, the collision frequency of the Xe atoms (with other atoms) e. the pressure of the Xe sample
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