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\(\bullet\) A large cylindrical tank contains 0.750 \(\mathrm{m}^{3}\) of nitrogen gas at \(27^{\circ} \mathrm{C}\) and \(1.50 \times 10^{5} \mathrm{Pa}\) (absolute pressure). The tank has a tight-fitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to 0.480 \(\mathrm{m}^{3}\) and the temperature is increased to \(157^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The final pressure is approximately \(3.377 \times 10^5 \text{ Pa}\).

Step by step solution

01

Convert temperature to Kelvin

First, we need to convert the temperatures from degrees Celsius to Kelvin. The formula to convert Celsius to Kelvin is: \[ T(K) = T(°C) + 273.15 \]So for the initial temperature:\[ T_1 = 27 + 273.15 = 300.15 \, \text{K} \]And for the final temperature:\[ T_2 = 157 + 273.15 = 430.15 \, \text{K} \]
02

Apply the ideal gas law relationship

Using the ideal gas law in terms of two states, we have:\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]We know:\( P_1 = 1.50 \times 10^5 \, \text{Pa} \), \( V_1 = 0.750 \, \text{m}^3 \), and \( T_1 = 300.15 \, \text{K} \)\( V_2 = 0.480 \, \text{m}^3 \) and \( T_2 = 430.15 \, \text{K} \).We need to solve for the final pressure \( P_2 \).
03

Solve for the final pressure

Rearrange the formula to solve for \( P_2 \):\[ P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \]Substitute the known values:\[ P_2 = \frac{(1.50 \times 10^5) \times 0.750 \times 430.15}{300.15 \times 0.480} \]Calculate the value:\[ P_2 \approx 3.377 \times 10^5 \, \text{Pa} \]
04

Conclusion

The pressure of the gas when the volume is decreased to \(0.480 \, \text{m}^3\) and the temperature is increased to \(157^{\circ} \, \text{C}\) is approximately \(3.377 \times 10^5 \, \text{Pa} \).

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Key Concepts

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

Pressure-Volume-Temperature Relationship
The pressure, volume, and temperature of a gas are interconnected in a fascinating way. This relationship is explained by the ideal gas law, which is a core concept in physics and chemistry. The law is typically expressed in the formula: \( PV = nRT \), where:
- \( P \) is the pressure of the gas,
- \( V \) is the volume,
- \( n \) is the number of moles of the gas,
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin.

In a closed system, where the amount of gas doesn't change, the ideal gas law can also be modified to compare two states of the same gas. It becomes: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). This equation helps us predict how changes in one condition (such as temperature or volume) will affect the others, assuming the amount of gas remains constant.
Understanding this relationship is vital in solving problems like the one presented in the exercise, where we want to find the new pressure after altering both temperature and volume while maintaining the amount of gas.
Temperature Conversion
Temperature is a crucial factor in gas behavior, but it’s vital to use the correct scale for calculations. The Kelvin scale is the standard unit used in gas equations because it starts at absolute zero (the point where particles have minimal vibrational motion), ensuring that all calculations remain mathematically consistent.

To convert Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. This is crucial because the Celsius scale doesn't align with the principles of thermodynamics used in calculations like those in gas laws:
  • Initial temperature: \( 27^{\circ} \text{C} \) converts to \( 300.15 \text{ K} \)

  • Final temperature: \( 157^{\circ} \text{C} \) converts to \( 430.15 \text{ K} \)

Using Kelvin ensures that our calculations within the gas laws reflect the true thermodynamic temperatures, allowing for accurate predictions of gas behavior.
Gas Laws
Gas laws collectively describe how gases behave under different conditions of pressure, volume, and temperature. The ideal gas law is a combination of several simpler laws that each describe how gases respond to a change in one variable while keeping others constant:

  • Boyle’s Law: Describes the inverse relationship between volume and pressure. As volume decreases, pressure increases (assuming constant temperature).

  • Charles’ Law: Explains that volume and temperature are directly proportional. As temperature increases, so does volume, if pressure remains constant.

  • Gay-Lussac’s Law: Shows that pressure and temperature are directly proportional. Increasing the temperature increases pressure, if volume remains constant.

The ideal gas law synthesizes these observations into a single expression, useful for analyzing real-world scenarios where more than one variable changes simultaneously. In the exercise provided, understanding these gas laws allows you to track how the compressed gas reacts to both changes in volume and increased temperature, resulting in a distinct change in pressure.

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Most popular questions from this chapter

\(\bullet\) A cylinder contains 0.250 mol of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas at a temperature of \(27.0^{\circ} \mathrm{C}\) . The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to \(127.0^{\circ} \mathrm{C}\) . Assume that the \(\mathrm{CO}_{2}\) may be treated as an ideal gas. (a) Draw a \(p V\) diagram of this process. (b) How much work is done by the gas in the process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

\(\bullet$$\bullet\) We have two equal-size boxes, \(\mathrm{A}\) and \(\mathrm{B}\) . Each box contains gas that behaves as an ideal gas. We insert a thermometer into each box and find that the gas in box \(A\) is at a temperature of \(50^{\circ} \mathrm{C}\) while the gas in box \(B\) is at \(10^{\circ} \mathrm{C} .\) This is all we know about the gas in the boxes. Which of the following statements must be true? Which could be true? (a) The pressure in \(A\) is higher than in \(B\) . (b) There are more molecules in \(A\) than in \(B\) . (c) \(A\) and \(B\) cannot contain the same type of gas. (d) The molecules in \(A\) have more average kinetic energy per molecule than those in \(B\) (e) The molecules in \(A\) are moving faster than those in \(B\) . Explain the reasoning behind your answers.

\(\bullet\) A gas under a constant pressure of \(1.50 \times 10^{5} \mathrm{Pa}\) and with an initial volume of 0.0900 \(\mathrm{m}^{3}\) is cooled until its volume becomes 0.0600 \(\mathrm{m}^{3} .\) (a) Draw a \(p V\) diagram of this process. (b) Calculate the work done by the gas.

\(\bullet\) An ideal gas expands while the pressure is kept constant. During this process, does heat flow into the gas or out of the gas? Justify your answer.

\(\bullet\) Five moles of an ideal monatomic gas with an initial temperature of \(127^{\circ} \mathrm{C}\) expand and, in the process, absorb 1200 \(\mathrm{J}\) of heat and do 2100 \(\mathrm{J}\) of work. What is the final temperature of the gas?

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