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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 "C. What is the volume of flask B?

Short Answer

Expert verified
The volume of flask B is approximately 117.34 mL.

Step by step solution

01

Identify Known Variables

From the problem, we know the initial volume \(V_1 = 25.0 \text{ mL}\), the initial pressure \(P_1 = 436.5 \text{ mm Hg}\), and the initial temperature \(T_1 = 20.5^{\circ} \text{C} = 293.65 \text{ K}\) (converted to Kelvin by adding 273.15). In flask B, the pressure \(P_2 = 94.3 \text{ mm Hg}\) and the temperature \(T_2 = 24.5^{\circ} \text{C} = 297.65 \text{ K}\). We need to find the volume \(V_2\) of flask B.
02

Use the Combined Gas Law

The Combined Gas Law is given by \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). We will use this equation to solve for \(V_2\), the unknown volume of flask B. Rearrange the formula to get \(V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}\).
03

Substitute Known Values

Substitute the known values into the equation: \[V_2 = \frac{(436.5 \text{ mm Hg}) \times (25.0 \text{ mL}) \times (297.65 \text{ K})}{(94.3 \text{ mm Hg}) \times (293.65 \text{ K})}\] Calculate this expression to find \(V_2\).
04

Calculate Final Volume

Compute the numerical value:\[V_2 = \frac{(436.5) \times (25.0) \times (297.65)}{(94.3) \times (293.65)} = \frac{3245878.25}{27658.645}\approx 117.34 \text{ mL}\] Thus, the volume of flask B is approximately 117.34 mL.

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Key Concepts

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

Gas Laws
Gas laws describe the behavior of gases, primarily focusing on their relationships between pressure, volume, and temperature. Among the many gas laws, the Combined Gas Law is particularly helpful when dealing with changes in a gas's condition involving more than one property. It combines the three main individual gas laws: Boyle's Law (pressure and volume relationship), Charles' Law (volume and temperature relationship), and Gay-Lussac's Law (pressure and temperature relationship).

The equation for the Combined Gas Law is \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). It implies that the state of a fixed amount of gas can be altered by changing its pressure, volume, and temperature, as long as the amount of gas remains constant.
  • The subscript '1' represents the initial state of the gas, and '2' represents the final state.
  • The pressure (P), volume (V), and temperature (T) must be measured using consistent units.
Understanding how these properties change in concert allows us to predict what would happen when a gas is subjected to different conditions, as demonstrated in our problem solving.
Temperature Conversion
Temperature conversion is a critical skill for solving problems involving gas laws because temperatures must be converted to Kelvin when applying these laws. This is due to the absolute nature of the Kelvin scale, which starts at absolute zero (-273.15°C), where theoretically, all molecular motion stops.

To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. In the exercise, the initial temperature was converted as follows:
  • For 20.5°C, the calculation is: \(20.5 + 273.15 = 293.65 \text{ K}\).
  • For 24.5°C, it becomes: \(24.5 + 273.15 = 297.65 \text{ K}\).
Using Kelvin ensures that we are using a temperature scale suitable for scientific calculations, particularly the ones involving gas law equations, maintaining a direct proportionality that Celsius lacks.
Pressure Measurement
Pressure is defined as the force exerted per unit area, and it's crucial in gas law calculations. In the context of gases, pressure results from gas molecules colliding with the walls of their container.

Pressure can be measured in various units, including millimeters of mercury (mm Hg), atmospheres (atm), and pascals (Pa). The exercise uses mm Hg, a common pressure unit in laboratory settings.
  • The initial pressure ( P_1 ) given was 436.5 mm Hg.
  • After transferring the gas to flask B, the pressure ( P_2 ) was 94.3 mm Hg.
Monitoring pressure changes is important because it directly affects the volume of gases, as illustrated by the Combined Gas Law. Understanding how pressure measures and converts is essential for accurately predicting state changes in gas.
Volume Calculation
Volume is the space that a gas occupies, and calculating changes in volume is essential in understanding gas behavior under different conditions. In our exercise, we needed to calculate the final volume (V_2) of flask B when the gas was transferred under different pressure and temperature conditions.

Using the Combined Gas Law, we solved for V_2: \( V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \). The known values were plugged in:
  • P_1 = 436.5 mm Hg, V_1 = 25.0 mL, T_1 = 293.65 K
  • P_2 = 94.3 mm Hg, T_2 = 297.65 K
After performing the calculation, the volume of flask B was found to be approximately 117.34 mL.

Accurate volume calculation is necessary because it allows us to determine how different conditions affect the amount of space gas occupies, helping in planning and performing actions like gas storage or reactions.

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Most popular questions from this chapter

The temperature of the atmosphere on Mars can be as high as \(27^{\circ} \mathrm{C}\) at the equator at noon, and the atmospheric pressure is about \(8 \mathrm{mm}\) Hg. If a spacecraft could collect \(10 . \mathrm{m}^{3}\) of this atmosphere, compress it to a small volume, and send it back to Earth, how many moles would the sample contain?

You have 3.6 L of \(\mathrm{H}_{2}\) gas at \(380 \mathrm{mm} \mathrm{Hg}\) and \(25^{\circ} \mathrm{C} .\) What is the pressure of this gas if it is transferred to a 5.0 -L flask at \(0.0^{\circ} \mathrm{C} ?\)

Under which set of conditions will \(\mathrm{Cl}_{2}\) deviate least from ideal gas behavior? (a) \(1 \mathrm{atm}, 0^{\circ} \mathrm{C}\) (b) 0.1 atm, \(100^{\circ} \mathrm{C}\) (c) \(10 \mathrm{atm}, 0^{\circ} \mathrm{C}\) (d) \(1 \mathrm{atm}, 100^{\circ} \mathrm{C}\)

A \(0.0130-\mathrm{g}\) sample of a gas with an empirical formula of \(\mathrm{C}_{4} \mathrm{H}_{5}\) is placed in a 165 -mL flask. It has a pressure of \(13.7 \mathrm{mm}\) Hg at \(22.5^{\circ} \mathrm{C}\). What is the molecular formula of the compound?

Iron carbonyl can be made by the direct reaction of iron metal and carbon monoxide. $$ \mathrm{Fe}(\mathrm{s})+5 \mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{Fe}(\mathrm{CO})_{5}(\ell) $$ What is the theoretical yield of \(\mathrm{Fe}(\mathrm{CO})_{5}\) if \(3.52 \mathrm{g}\) of iron is treated with CO gas having a pressure of \(732 \mathrm{mm} \mathrm{Hg}\) in a \(5.50-\mathrm{L}\) flask at \(23^{\circ} \mathrm{C} ?\)

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