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

Short answer

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

Step-by-step solution

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.

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, \(T_2 = 88°C + 273.15 = 361.15 K\)

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, \(V_1\). \(V_1 = \frac{4}{3} \pi (1.00)^3 = \frac{4}{3} \pi cm^3\)

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: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)

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: \(V_2 = \frac{4}{3} \pi cm^3 \times \frac{361.15}{280.15} \approx 1.89 \pi cm^3\)

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.