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Chapter 10: Problem 56

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

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

Step by step solution

01

Convert the temperatures to Kelvin

To convert temperatures from Celsius to Kelvin, add 273.15 to the Celsius temperature. Therefore, we have: \(T_i = 45^{\circ} \mathrm{C} + 273.15 = 318.15 \, \mathrm{K}\) \(T_f = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \, \mathrm{K}\)
02

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.
03

Calculate the change in volume

We can express the change in volume as: \(V_f - V_i = k * T_f - k * T_i = k * (T_f - T_i)\)
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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掳C, the conversion would be:
  • First, take the Celsius temperature, 45掳C.
  • Then add 273.15 to it, resulting in 318.15 K.
Similarly, if the temperature drops to 25掳C, the conversion goes like this:
  • Add 273.15 to 25掳C.
  • This gives 298.15 K.
Using Kelvin in computations ensures accuracy when applying formulas such as Charles鈥檚 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.

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Most popular questions from this chapter

Using Wetlands to Treat Agricultural Waste Wetlands can play a significant role in removing fertilizer residues from rain runoff and groundwater. One way they do this is through denitrification, which converts nitrate ions to nitrogen gas: \(2 \mathrm{NO}_{3}^{-}(a q)+5 \mathrm{CO}(g)+2 \mathrm{H}^{+}(a q) \right-arrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)+5 \mathrm{CO}_{2}(g)\) Suppose \(200.0 \mathrm{g}\) of \(\mathrm{NO}_{3}^{-}\) flows into a swamp each day. a. What volume of \(\mathrm{N}_{2}\) would be produced at \(17^{\circ} \mathrm{C}\) and 1.00 atm if the denitrification process were complete? b. What volume of \(\mathrm{CO}_{2}\) would be produced? c. Suppose the gas mixture produced by the decomposition reaction is trapped in a container at \(17^{\circ} \mathrm{C} ;\) what is the density of the mixture, assuming \(P_{\text {total }}\) \(=1.00\) atm?

Decay Products of Uranium Minerals Radon and helium are both by-products of the radioactive decay of uranium minerals. A fresh sample of carnotite, \(\mathrm{K}_{2}\left(\mathrm{UO}_{2}\right)_{2}\left(\mathrm{VO}_{4}\right)_{2}\) \(\cdot 3 \mathrm{H}_{2} \mathrm{O},\) is put on display in a museum. Calculate the relative rates of diffusion of helium and radon under fixed conditions of pressure and temperature. Which gas diffuses more rapidly through the display case?

The following reactions are carried out in sealed containers. Will the total pressure after each reaction is complete be greater than, less than, or equal to the total pressure before the reaction? Assume all reactants and products are gases at the same temperature. a. \(\mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{NO}_{2}(g) \rightarrow 3 \mathrm{NO}(g)+2 \mathrm{O}_{2}(g)\) b. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\) c. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

Derive an equation that expresses the ratio of the densities \(\left(d_{1} \text { and } d_{2}\right)\) of a gas under two different combinations of temperature and pressure: \(\left(T_{1}, P_{1}\right)\) and \(\left(T_{2}, P_{2}\right)\).

A balloon vendor at a street fair is using a tank of helium to fill her balloons. The tank has an internal volume of \(45.0 \mathrm{L}\) and a pressure of 195 atm at \(22^{\circ} \mathrm{C}\). After a while she notices that the valve has not been closed properly and the pressure has dropped to 115 atm. How many moles of He have been lost?
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