Problem 5 - Arihant AIEEE Physics

Question

\(12 \mathrm{~g}\) of gas occupy a volume of \(4 \times 10^{-3} \mathrm{~m}^{3}\) at a temperature of \(7^{\circ} \mathrm{C}\). After the gas is heated at constant pressure, its density becomes \(6 \times 10^{-4} \mathrm{~g} / \mathrm{cm}^{3}\). What is the temperature to which the gas was heated? (a) \(1000 \mathrm{~K}\) (b) \(1400 \mathrm{~K}\) (c) \(1200 \mathrm{~K}\) (d) \(800 \mathrm{~K}\)

Step-by-Step Solution

Answer

The final temperature is 1400 K (option b).

Step 1: Analyze Initial Conditions

We are given that the gas initially has a mass of 12 g, a volume of \(4 \times 10^{-3} \text{ m}^3\), and a temperature of \(7^\circ \text{C}\). This temperature can be converted to Kelvin: \(T_1 = 7 + 273.15 = 280.15 \text{ K}\).

Step 2: Determine Initial Density

The initial density \( \rho_1 \) is calculated using the formula \( \rho_1 = \frac{\text{mass}}{\text{volume}} \). Substituting the given values, \( \rho_1 = \frac{12 \text{ g}}{4 \times 10^{-3} \text{ m}^3} = 3000 \text{ g/m}^3 = 3 \text{ g/cm}^3\).

Step 3: Apply Ideal Gas Law to Find Final Volume

At constant pressure, \[P_i V_i = nRT_i\] and \[P_f V_f = nRT_f\] Given that the pressure is constant, \(\frac{V_i}{T_i} = \frac{V_f}{T_f}\). This implies that \(V_f = V_i \frac{T_f}{T_i}\).

Step 4: Calculate Final Volume Using Change in Density

We know the final density \( \rho_2 = 6 \times 10^{-4} \text{ g/cm}^3 \), which needs to be converted to g/m³: \(\rho_2 = 0.6 \text{ g/m}^3\). Using \(\rho_2 = \frac{12}{V_f}\), solve for \(V_f\): \(V_f = \frac{12}{0.6} = 20 \text{ m}^3\).

Step 5: Apply Constant Pressure Condition and Calculate Final Temperature

Using the proportionality at constant pressure, \(\frac{V_f}{V_i} = \frac{T_f}{T_i}\), isolate for \(T_f\): \(T_f = \frac{V_f}{V_i} T_i\). Substitute the values: \( V_i = 0.004 \text{ m}^3, V_f = 20 \text{ m}^3, T_i = 280.15 \text{ K} \)\[T_f = \frac{20}{0.004} \times 280.15 = 1400 \text{ K} \].

Key Concepts

Density Calculation

Density plays a crucial role in understanding how gas behaves when it undergoes temperature and volume changes. In the context of gases, density is defined as the mass per unit volume. The formula for calculating density \( \rho \) is:\[\rho = \frac{\text{mass}}{\text{volume}}\]For example, in the given exercise, the initial density \( \rho_1 \) was calculated based on a mass of 12 grams and an initial volume of \(4 \times 10^{-3} \, \text{m}^3\). Here’s how it’s done:

• Density \( \rho_1 = \frac{12}{4 \times 10^{-3}} \text{ g/m}^3 = 3000 \text{ g/m}^3 \linebreak = 3 \text{ g/cm}^3 \)

By knowing the initial density, we can determine how changes like temperature increase can impact the gas volume and density again later. In this exercise, after the gas is heated, its density changes. Calculating the density helps in solving for the new conditions of the gas.

Gas Volume Change

Gas volume is sensitive to changes in temperature and pressure, as dictated by the Ideal Gas Law and its derived relationships. When a gas is kept at constant pressure, as in this exercise, changes in temperature result in changes in volume. The relationship at constant pressure can be represented as:\[\frac{V_i}{T_i} = \frac{V_f}{T_f}\]where:

• \( V_i \) is the initial volume
• \( V_f \) is the final volume
• \( T_i \) is the initial temperature in Kelvin
• \( T_f \) is the final temperature in Kelvin

In the exercise, the initial gas volume was given as \(4 \times 10^{-3} \, \text{m}^3\). After heating, the calculated final volume was found to be 20 \(\text{m}^3\). This substantial change illustrates the responsiveness of gases to heating under constant pressure, where volume increases significantly as temperature rises.

Temperature Conversion

Converting temperatures is vital when working with gases, primarily because most gas laws use absolute temperature scales like Kelvin. To convert Celsius to Kelvin, add 273.15 to the Celsius temperature:\[T(K) = T(^\circ C) + 273.15\]In the original problem, the initial temperature of the gas was given as \(7^\circ \text{C}\). Changing it to Kelvin simplifies solving the equations and ensures all calculations reflect absolute thermal conditions. Therefore,

• \(T_i = 7 + 273.15 = 280.15 \, \text{K}\)

Using Kelvin helps avoid negative temperatures which can occur in Celsius, thus maintaining consistency when applying gas laws like Boyle's or Charles’s Law.

Constant Pressure Conditions

Understanding constant pressure conditions is key when analyzing how gas properties change. Under constant pressure, the Ideal Gas Law simplifies to examine volume and temperature relationships.

• At constant pressure, \( P_i V_i = nRT_i \) and \( P_f V_f = nRT_f \)
• This leads to the proportionality \( \frac{V_i}{T_i} = \frac{V_f}{T_f} \)

This proportionality suggests that, as temperature increases, volume increases in direct proportion under constant pressure. For the exercise, employing this relationship clarified how the volume expanded from an initial \(4 \times 10^{-3} \, \text{m}^3\) to a final volume of 20 \(\text{m}^3\). This principle allowed for the calculation of the new temperature \( T_f \) to be correctly deduced, reflecting how temperature is directly linked to volume when pressure remains steady.