Question

The air inside a balloon is heated to \(45^{\circ} \mathrm{C}\) and then cools to \(25^{\circ} \mathrm{C} .\) By what percentage does the volume of the balloon change during cooling?

Short answer

Answer: The percentage change in the volume of the balloon when it cools from 45°C to 25°C is approximately -6.28%. This means that the volume of the balloon decreases by 6.28%.

Step-by-step solution

Convert the temperatures to Kelvin

To convert temperatures from Celsius to Kelvin, add 273.15 to the Celsius temperature. Therefore, we have: \(T_i = 45^{\circ} \mathrm{C} + 273.15 = 318.15 \, \mathrm{K}\) \(T_f = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \, \mathrm{K}\)

Write the general formulas for the initial and final states

Assuming that pressure (P) and the amount of air (n) inside the balloon remain constant, we can use the equation PV = nRT (where R is the gas constant) to write the formulas for the initial and final states: Initial state: \(V_i = k * T_i\) Final state: \(V_f = k * T_f\) Where k is a constant.

Calculate the change in volume

We can express the change in volume as: \(V_f - V_i = k * T_f - k * T_i = k * (T_f - T_i)\)

Calculate the percentage change in volume

Using the change in volume (from Step 3), we can now calculate the percentage change in volume: Percentage change in volume = \(\frac{V_f - V_i}{V_i} * 100 = \frac{k*(T_f - T_i)}{k*T_i} * 100\) Substitute the values of \(T_i, T_f\) which we found in Step 1: Percentage change in volume = \(\frac{(298.15 - 318.15)}{318.15} * 100 = -6.28\%\) The negative sign indicates that the volume of the balloon decreases by 6.28% during the cooling process.

Key concepts

Temperature Conversion

Understanding temperature conversion is key when dealing with gas laws. Specifically, the Kelvin scale is used in scientific calculations because it starts at absolute zero, making it useful for gas-related problems. To convert temperatures from Celsius to Kelvin, simply add 273.15 to the Celsius temperature.

For instance, if a balloon's air is heated to 45°C, the conversion would be:

• First, take the Celsius temperature, 45°C.
• Then add 273.15 to it, resulting in 318.15 K.

Similarly, if the temperature drops to 25°C, the conversion goes like this:

• Add 273.15 to 25°C.
• This gives 298.15 K.

Using Kelvin in computations ensures accuracy when applying formulas such as Charles’s Law.

Charles's Law

Charles's Law describes how gases tend to expand when heated. It directly relates the temperature of a gas to its volume, under constant pressure and amount of gas.

• The formula is given as \( V \propto T \) or \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( V \) is volume and \( T \) is temperature in Kelvin.

This means that if the temperature of a gas increases, its volume increases as well, assuming pressure and quantity of gas remain unchanged. In our exercise, initially, when the balloon is heated to 318.15 K, the volume is proportionately larger compared to when it cools down to 298.15 K. In both scenarios, using Kelvin ensures accurate assessment of how volume changes with temperature.

Volume Change Calculation

Calculating the change in volume as a percentage involves using initial and final temperatures to determine how much the volume shifts due to temperature changes.

Given the exercise, we determine this change as follows:

• First, understand that \( V_i = k * T_i \) and \( V_f = k * T_f \), where \( k \) is a constant based on initial conditions.
• Next, express the volume change using the formula: \( V_f - V_i = k * (T_f - T_i) \).

Now, calculating the percentage volume change:

• Plug in the values for \( T_i = 318.15 \) and \( T_f = 298.15 \).
• Percentage change is calculated as \( \frac{V_f - V_i}{V_i} * 100 \) which simplifies to \( \frac{k*(298.15 - 318.15)}{k*318.15} * 100 \).

When simplified, this shows a decrease of 6.28%, indicating a contraction in the balloon's volume with cooling. This calculation illustrates how even small temperature changes can impact gas volume.