/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 You have a sample of \(\mathrm{C... [FREE SOLUTION] | 91Ó°ÊÓ

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You have a sample of \(\mathrm{CO}_{2}\) in flask \(\mathrm{A}\) with a volume of \(25.0 \mathrm{mL} .\) At \(20.5^{\circ} \mathrm{C},\) the pressure of the gas is \(436.5 \mathrm{mm}\) Hg. To find the volume of another flask, B, you move the \(\mathrm{CO}_{2}\) to that flask and find that its pressure is now \(94.3 \mathrm{mm}\) Hg at \(24.5^{\circ} \mathrm{C} .\) What is the volume of flask \(\mathrm{B} ?\)

Short Answer

Expert verified
The volume of flask B is approximately 117.13 mL.

Step by step solution

01

Identify the Known Variables

We have the initial conditions for flask A: \(V_1 = 25.0 \mathrm{mL}\), \(P_1 = 436.5 \mathrm{mmHg}\), and \(T_1 = 20.5^{\circ}C\). Convert the temperature to Kelvin: \(T_1 = 20.5 + 273.15 = 293.65 \text{ K}\). We also have the final conditions for flask B: \(P_2 = 94.3 \mathrm{mmHg}\) and \(T_2 = 24.5^{\circ}C\). Convert this temperature to Kelvin: \(T_2 = 24.5 + 273.15 = 297.65 \text{ K}\).
02

Apply the Combined Gas Law

The combined gas law can be applied because the number of moles of gas remains constant: \[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\].We need to solve for \(V_2\), the volume of flask B, so rearrange the equation to find: \[V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}\].
03

Insert the Known Values and Solve

Insert the known values into the rearranged formula:\[V_2 = \frac{436.5 \mathrm{mmHg} \times 25.0 \mathrm{mL} \times 297.65 \text{ K}}{94.3 \mathrm{mmHg} \times 293.65 \text{ K}}\].Calculate each part:\[V_2 = \frac{3241264.625 \text{ mL} \cdot \mathrm{K}}{27676.395 \text{ K \cdot mmHg}}\].Simplifying gives:\[V_2 \approx 117.13 \mathrm{mL}\].
04

Conclusion: Determine the Final Volume

The volume of flask B is therefore approximately \(117.13 \mathrm{mL}\).

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Key Concepts

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

Gas Pressure
Gas pressure is a crucial property in understanding how gases behave. It refers to the force exerted by gas molecules when they collide with the walls of their container. The more frequently and forcefully they collide, the higher the pressure.
Factors affecting gas pressure include:
  • The number of gas molecules. More molecules mean more collisions, resulting in higher pressure.
  • The temperature of the gas. A higher temperature causes molecules to move faster, increasing collision frequency and pressure.
  • The volume of the container. A smaller volume means gas molecules are closer together, leading to more collisions.
In the exercise, the pressure of the carbon dioxide (\(\mathrm{CO}_{2}\)) gas in flask A starts at 436.5 mm Hg, but decreases to 94.3 mm Hg in flask B. This change demonstrates how pressure can vary with changes in the environment.
Gas Volume
Gas volume is the space that a gas occupies, and it's subject to change based on pressure and temperature. In general, gases expand to fill their containers and adjust their volume based on surroundings.
When the gas is transferred from flask A to flask B, its volume changes because of the different pressure and temperature conditions. To calculate the new volume using the combined gas law, we keep the amount of gas constant and relate the initial and final conditions of pressure, volume, and temperature. In the formula, volume is adjusted according to pressure levels and temperature changes. The rule of thumb is:
  • Lower pressure allows a gas to occupy a larger volume.
  • Higher pressure compresses the gas, reducing its volume.
In the exercise, flask B's volume increased to approximately 117.13 mL due to a decrease in pressure while some increase in temperature was accounted for.
Temperature Conversion
Temperature conversion is an essential step in gas law calculations as temperatures must be in Kelvin. The Kelvin scale starts from absolute zero, making it ideal for scientific calculations.
To convert temperature from Celsius to Kelvin, use the formula: \[K = °C + 273.15\]This conversion ensures accurate calculation with gas laws since Kelvin provides an absolute measure of thermal energy.
In the exercise, temperatures are converted from Celsius as follows:
  • Flask A: \(T_1 = 20.5^{\circ}\, C \to T_1 = 293.65 \, K\)
  • Flask B: \(T_2 = 24.5^{\circ}\, C \to T_2 = 297.65 \, K\)
This step guarantees that changes in thermal conditions are accounted for in solving for the gas’s volume.

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Most popular questions from this chapter

A sample of \(\mathrm{CO}_{2}\) gas has a pressure of \(56.5 \mathrm{mm} \mathrm{Hg}\) in a 125-mi. flask. The sample is transferred to a new flask, where it has a pressure of \(62.3 \mathrm{mm}\) Hg at the same temperature. What is the volume of the new flask?

Chlorine trifluoride, \(\mathrm{CIF}_{3}\), is a valuable reagent because it can be used to convert metal oxides to metal fluorides: \(6 \mathrm{NiO}(\mathrm{s})+4 \mathrm{ClF}_{3}(\mathrm{g}) \rightarrow 6 \mathrm{NiF}_{2}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g})\) (a) What mass of NiO will react with CIF \(_{3}\) gas if the gas has a pressure of \(250 \mathrm{mm} \mathrm{Hg}\) at \(20^{\circ} \mathrm{C}\) in a \(2.5-\mathrm{L}\) flask? (b) If the CIF a described in part (a) is completely consumed, what are the partial pressures of \(\mathrm{Cl}_{2}\) and of \(\mathrm{O}_{2}\) in the 2.5 -I. flask at \(20^{\circ} \mathrm{C}\) (in \(\mathrm{mm}\) Hg)? What is the total pressure in the flask?

Argon gas is 10 times denser than helium gas at the same temperature and pressure. Which gas is predicted to effuse faster? How much faster?

The ideal gas law is least accurate under conditions of high pressure and low temperature. In those situations, using the van der Waals equation is advisable. (a) Calculate the pressure exerted by \(12.0 \mathrm{g}\) of \(\mathrm{CO}_{2}\) in a \(500-\mathrm{mL}\) vessel at \(298 \mathrm{K},\) using the ideal gas equation. Then, recalculate the pressure using the van der Waals equation. Assuming the pressure calculated from van der Waal's equation is correct, what is the percent error in the answer when using the ideal gas equation? (b) Next, cool this sample to \(-70^{\circ} \mathrm{C}\). Then perform the same calculation for the pressure exerted by \(\mathrm{CO}_{2}\) at this new temperature, using both the ideal gas law and the van der Waals equation. Again, what is the percent error when using the ideal gas equation?

Ethane burns in air to give \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\) $$2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ What volume of \(\mathrm{O}_{2}\) (L) is required for complete reaction with \(5.2 \mathrm{L}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} ?\) What volume of \(\mathrm{H}_{2} \mathrm{O}\) vapor ( \(\mathrm{L}\) ) is produced? Assume all gases are measured at the same temperature and pressure.

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