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Chapter 13: Problem 63

What will the volume of the sample become if \(459 \mathrm{~mL}\) of an ideal gas at \(27^{\circ} \mathrm{C}\) and 1.05 atm is cooled to 15 C and 0.997 atm?

Short Answer

Expert verified
The final volume of the sample under the given conditions will be approximately \(473.74 \mathrm{~mL}\).

Step by step solution

01

Convert temperatures to Kelvin

To work with the combined gas law, we need to convert the given temperatures from Celsius to Kelvin. We can do this using the following formula: \( T_K = T_C + 273.15 \) Initial temperature in Kelvin (T1): \( T_1 = 27^{\circ} \mathrm{C} + 273.15 = 300.15 \mathrm{K} \) Final temperature in Kelvin (T2): \( T_2 = 15^{\circ} \mathrm{C} + 273.15 = 288.15 \mathrm{K} \)
02

Plug in given values into the combined gas law

Now that we have the temperature values in Kelvin, we can plug all of the given values into the combined gas law formula: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) Where: - \(P_1 = 1.05 \mathrm{~atm}\) - \(V_1 = 459 \mathrm{~mL}\) - \(T_1 = 300.15 \mathrm{~K}\) - \(P_2 = 0.997 \mathrm{~atm}\) - \(T_2 = 288.15 \mathrm{~K}\) - \(V_2\) is the final volume we need to find
03

Rearrange equation and solve for final volume

To find \(V_2\), we first need to rearrange the combined gas law formula to isolate \(V_2\) on one side: \( V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \) Now, we can plug in the given values and solve for the final volume: \( V_2 = \frac{(1.05 \mathrm{~atm})(459 \mathrm{~mL})(288.15 \mathrm{~K})}{(0.997 \mathrm{~atm})(300.15 \mathrm{~K})} \)
04

Calculate the final volume

Calculate the final volume by performing the arithmetic: \( V_2 = \frac{(1.05)(459)(288.15)}{(0.997)(300.15)} \approx 473.74 \mathrm{~mL} \) Therefore, the final volume of the sample under the given conditions will be approximately 473.74 mL.

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Key Concepts

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

Temperature Conversion
To solve many gas law problems, converting temperatures from Celsius to Kelvin is a necessary first step. The Kelvin scale is the standard unit of measurement for thermodynamic temperature in scientific calculations. This is because the Kelvin scale starts at absolute zero, where molecular motion stops, providing a more natural reference point for mathematical computations.
The conversion method is straightforward: simply add 273.15 to the Celsius temperature.
  • For example, to convert 27掳C to Kelvin: \[ T_K = 27 + 273.15 = 300.15 \]
  • Similarly, for 15掳C: \[ T_K = 15 + 273.15 = 288.15 \]
Ensuring temperatures are in Kelvin when applying gas laws helps maintain uniformity and accuracy in your calculations.
Ideal Gas Law
The ideal gas law is a cornerstone of chemistry for understanding relationships among pressure, volume, temperature, and amount of gas. Specifically, the combined gas law, used in our problem, integrates aspects of the ideal gas law to cover scenarios where the amount of gas remains unchanged.
In the combined form, the law is represented as:\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]This equation connects initial and final states of gas, considering
  • \(P_1\) and \(P_2\) as initial and final pressures, respectively
  • \(V_1\) and \(V_2\) as initial and final volumes
  • \(T_1\) and \(T_2\) as initial and final temperatures in Kelvin
Using this formula allows you to predict how one parameter changes when others are altered, assuming constant gas quantity.
Arithmetic Calculations
After rearranging the combined gas law equation to solve for the final volume, perform arithmetic calculations carefully to ensure accuracy in your answer.The rearranged equation becomes:\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]To find \(V_2\), substitute the known values into the equation:\[ V_2 = \frac{(1.05)(459)(288.15)}{(0.997)(300.15)} \]Breaking down the calculation:
  • First, multiply the numerator: \(1.05 \times 459 \times 288.15 \)
  • Then, multiply the denominator: \(0.997 \times 300.15\)
  • Finally, divide the results of those calculations to find \(V_2\)
Executing each calculation in sequence will yield the final volume as approximately 473.74 mL. Proper arithmetic ensures the integrity of your solution.

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Most popular questions from this chapter

What is a scientific law? What is a theory? How do these concepts differ? Does a law explain a theory, or does a theory attempt to explain a law?

A sample of oxygen gas is saturated with water vapor at 27 . C. The total pressure of the mixture is 772 torr, and the vapor pressure of water is 26.7 torr at 27 . \(\mathrm{C}\). What is the partial pressure of the oxygen gas?

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For each of the following sets of volume/temperature data, calculate the missing quantity. Assume that the pressure and the mass of gas remain constant. a. \(V=25.0 \mathrm{~L}\) at \(0 \quad \mathrm{C} ; V=50.0 \mathrm{~L}\) at \(? \mathrm{C}\) b. \(V=247 \mathrm{~mL}\) at 25 ' \(\mathrm{C} ; V=255 \mathrm{~mL}\) at \(? \mathrm{C}\) c. \(V=1.00 \mathrm{~mL}\) at \(2272 \mathrm{C} ; V=?\) at \(25 \mathrm{C}\)

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