/*! 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 75 A gas sample at \(22^{\circ} \ma... [FREE SOLUTION] | 91Ó°ÊÓ

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A gas sample at \(22^{\circ} \mathrm{C}\) and 740 torr pressure is heated until its volume is doubled. What pressure would restore the sample to its original volume?

Short Answer

Expert verified
The pressure that would restore the gas sample to its original volume after doubling its volume at constant temperature is 370 torr.

Step by step solution

01

Convert temperature to Kelvin

Convert the given temperature from Celsius to Kelvin by adding 273.15 to it. The temperature in Kelvin is necessary for gas law calculations. Given temperature is 22 °C, so in Kelvin: Temperature(K) = 22 + 273.15 = 295.15 K.
02

Apply Charles's Law for the volume change

Charles's Law states that at constant pressure, the volume of a gas is directly proportional to its temperature. When the volume of the gas is doubled, the temperature must also double for constant pressure. However, the problem implies a constant temperature rather than pressure. We need to maintain the volume so we find the new doubled volume condition under constant temperature, which wouldn't change the pressure.
03

Apply Boyle's Law to find the new pressure

Boyle's Law states that at constant temperature, the product of the volume and pressure of an ideal gas is constant. Let the initial volume be V, initial pressure be 740 torr. After heating, the volume is 2V. To return to volume V, Boyle's Law gives us P1 * V1 = P2 * V2, where P2 is the pressure we're trying to find. So, 740 torr * V = P2 * (2V).
04

Solve for the new pressure P2

Divide both sides of the equation by 2V to solve for P2: P2 = (740 torr * V) / (2V) = 740 torr / 2 = 370 torr. Therefore, the pressure that would restore the sample to its original volume is 370 torr.

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Key Concepts

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

Gas Laws Chemistry
Gas laws are foundational principles in chemistry that describe the behavior of gases under various conditions of temperature, volume, and pressure. Boyle's Law and Charles's Law are two critical components of these gas laws.

Understanding Boyle's Law

Boyle's Law asserts that for a certain amount of gas at constant temperature, the volume is inversely proportional to the pressure. In simple terms, if you increase the pressure on a gas while keeping its temperature the same, the volume of the gas will decrease. Conversely, if the pressure decreases, the volume increases. Mathematically, it's expressed as: \( P_1V_1 = P_2V_2 \) where \(P\) is pressure, \(V\) is volume, and subscripts 1 and 2 represent the initial and final states of the gas, respectively.

Navigating Charles's Law

On the other hand, Charles's Law states that the volume of a gas is directly proportional to its absolute temperature when the pressure is held constant. If you heat a gas, its volume will expand if the pressure is kept unchanged. Cool it down, and the volume shrinks. This relationship is captured by the formula: \( V_1/T_1 = V_2/T_2 \) Here, \(T\) represents the temperature in Kelvin, and the subscripts indicate different conditions. Both laws are part of the ideal gas law, which combines the relationships between the temperature, volume, and pressure of a gas. These fundamental principles are critical in understanding the behavior of gases, and they apply to many real-world scenarios, such as breathing, tire inflation, and weather patterns.
Pressure and Volume Relationship
Let's delve deeper into how pressure and volume interact, a concept integral to Boyle's Law. Imagine you have a sealed syringe containing a fixed amount of gas with the plunger pulled out. When you push the plunger, you're exerting more force on the gas, effectively increasing its pressure. Since the amount of gas and the temperature remain constant, the only way to accommodate this change is to reduce the volume of the gas - it gets compressed.

Now, from the equation of Boyle's Law \(P_1V_1 = P_2V_2\), notice that the product of the initial pressure and volume is equal to the product of the final pressure and volume. This relationship indicates a seesaw effect; as one goes up, the other must come down. When applied to real problems, this law enables us to predict what happens to one variable when we manipulate the other while keeping the temperature constant.

  • If the volume doubles, the pressure must halve.
  • If the volume is reduced by a third, the pressure triples.
Understanding this concept allows students to solve various practical and theoretical problems in chemistry, physics, and engineering.
Temperature Conversion Celsius to Kelvin
Temperature scales can be a source of confusion, but the conversion from Celsius to Kelvin is a crucial skill in chemistry. The Kelvin scale is an absolute temperature scale commonly used in scientific calculations because it starts at absolute zero, the theoretical point where particles have minimal thermal motion.

Converting Celsius to Kelvin

To convert temperatures from Celsius to Kelvin, one simply adds 273.15 to the Celsius value. This offset accounts for the difference between the starting points of the two scales: 0°C is equivalent to 273.15K. The formula is as follows: \(T(K) = T(°C) + 273.15\)

For example, the boiling point of water is \(100°C\), which is \(100 + 273.15 = 373.15K\) in Kelvin. Similarly, room temperature, around \(22°C\), converts to \(22 + 273.15 = 295.15K\). This temperature conversion is essential for gas law calculations, as it ensures that temperature is expressed in absolute terms, allowing for accurate and consistent calculations.

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Most popular questions from this chapter

A soccer ball of constant volume \(2.24 \mathrm{~L}\) is pumped up with air to a gauge pressure of \(13 \mathrm{lb} /\) in. \({ }^{2}\) at \(20.0^{\circ} \mathrm{C}\). The molar mass of air is about \(29 \mathrm{~g} / \mathrm{mol}\). (a) How many moles of air are in the ball? (b) What mass of air is in the ball? (c) During the game, the temperature rises to \(30.0^{\circ} \mathrm{C}\). What mass of air must be allowed to escape to bring the gauge pressure back to its original value?

You are responsible for ensuring that the giant American eagle balloon stays inflated at the local Veterans Day parade. You inflate the balloon to a pressure of 976 torr using \(5.27 \times 10^{5} \mathrm{~mol}\) of helium in the morning when the temperature is \(12^{\circ} \mathrm{C}\). At the end of the day the temperature increases to \(31^{\circ} \mathrm{C}\) and \(15.0 \%\) of the helium seeps out of the balloon. (a) How many moles of air will be left in the balloon at the end of the day after \(15.0 \%\) is lost? (b) What will the pressure of the balloon be at the end of the day if the volume is unchanged?

When glucose is burned in a closed container, carbon dioxide gas and water are produced according to the following equation: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ How many liters of \(\mathrm{CO}_{2}\) at STP can be produced when \(1.50 \mathrm{~kg}\) of glucose are burned?

Three gases were added to the same 2.0-I. container. The total pressure of the gases was 790 torr at room temperature \(\left(25.0^{\circ} \mathrm{C}\right)\). If the mixture contained \(0.65 \mathrm{~g}\) of oxygen gas, \(0.58 \mathrm{~g}\) of carbon dioxide, and an unknown amount of nitrogen gas, determine the following: (a) the total number of moles of gas in the container (b) the number of grams of nitrogen in the container (c) the partial pressure of each gas in the mixture

Perform the following pressure conversions: (a) Convert 953 torr to \(\mathrm{kPa}\). (b) Convert \(2.98 \mathrm{kPa}\) to atm. (c) Convert \(2.77\) atm to \(\mathrm{mm} \mathrm{Hg}\). (d) Convert 372 torr to atm. (e) Convert \(2.81 \mathrm{~atm}\) to \(\mathrm{cm} \mathrm{Hg}\).

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