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Will the volume of a gas increase, decrease, or remain unchanged with each of the following sets of changes? (a) The pressure is decreased from 2 atm to 1 atm, while the temperature is decreased from \(200^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C} .\) (b) The pressure is increased from 1 atm to 3 atm, while the temperature is increased from \(100^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\). (c) The pressure is increased from 3 atm to 6 atm, while the temperature is increased from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\). (d) The pressure is increased from 0.2 atm to 0.4 atm, while the temperature is decreased from \(300^{\circ} \mathrm{C}\) to \(150^{\circ} \mathrm{C}\).

Short Answer

Expert verified
(a) Increase(b) Decrease(c) Remain Unchanged(d) Decrease

Step by step solution

01

Understanding the Ideal Gas Law

The Ideal Gas Law is given by the equation: \[ PV = nRT \]where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature in Kelvins. From this, the volume V can be expressed as:\[ V = \frac{nRT}{P} \].This means that volume is directly proportional to temperature and inversely proportional to pressure.
02

Converting Temperatures to Kelvin

Temperatures given in Celsius need to be converted to Kelvin. The conversion formula is: \[ T(K) = T(^{\circ}C) + 273.15 \](a) Convert: \(200^{\circ}C\) to Kelvin: \(200 + 273.15 = 473.15 \text{ K}\).\(100^{\circ}C\) to Kelvin: \(100 + 273.15 = 373.15 \text{ K}\).(b) Convert: \(100^{\circ}C\) to Kelvin: \(100 + 273.15 = 373.15 \text{ K}\).\(300^{\circ}C\) to Kelvin: \(300 + 273.15 = 573.15 \text{ K}\).(c) Convert: \(-73^{\circ}C\) to Kelvin: \(-73 + 273.15 = 200.15 \text{ K}\).\(127^{\circ}C\) to Kelvin: \(127 + 273.15 = 400.15 \text{ K}\).(d) Convert: \(300^{\circ}C\) to Kelvin: \(300 + 273.15 = 573.15 \text{ K}\).\(150^{\circ}C\) to Kelvin: \(150 + 273.15 = 423.15 \text{ K}\).
03

Analyzing the Volume Change for Each Case

(a) The volume change is influenced by both the pressure and the temperature changes. Initially, the volume is:\[ V_1 = \frac{nR(473.15)}{2} \].Final volume is:\[ V_2 = \frac{nR(373.15)}{1} \].Since both the pressure is reduced by half and temperature is decreased (by less than half), the net effect is an increase in volume.(b) Initial volume:\[ V_1 = \frac{nR(373.15)}{1} \].Final volume:\[ V_2 = \frac{nR(573.15)}{3} \].Even though the temperature is almost doubled, the pressure is tripled, hence the net effect is a decrease in volume.(c) Initial volume:\[ V_1 = \frac{nR(200.15)}{3} \].Final volume:\[ V_2 = \frac{nR(400.15)}{6} \].Both the pressure and temperature are doubled, resulting in a net effect of unchanged volume.(d) Initial volume:\[ V_1 = \frac{nR(573.15)}{0.2} \].Final volume:\[ V_2 = \frac{nR(423.15)}{0.4} \].The pressure is doubled while the temperature is reduced (by less than half), leading to a net decrease in volume.

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Key Concepts

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

Ideal Gas Law
When studying gases, it's essential to understand the Ideal Gas Law, which is expressed through the formula: ewlineewline\[ PV = nRT \]ewlineewlineHere, P represents the pressure of the gas, V symbolizes volume, n denotes the number of moles, R is the universal gas constant, and T signifies the temperature in Kelvin. ewlineThis equation reveals that the volume (V) of a gas is directly related to its temperature (T) and inversely related to its pressure (P). This means: ewline\[ V = \frac{nRT}{P} \]ewlineewlineSo, if the temperature goes up, the volume increases—with a constant gas amount and pressure. Conversely, if the pressure increases, the volume decreases, assuming temperature and gas amount remain constant. ewlineUnderstanding this relationship helps in predicting how a gas behaves under varying conditions of temperature and pressure.
Pressure-Volume Relationship
The pressure-volume relationship is a critical component of gas behavior. It states that for a given amount of gas at constant temperature, the volume of the gas varies inversely with its pressure. This is known as Boyle's Law. ewlineIn formula terms, it is expressed as: ewline\[ P_1V_1 = P_2V_2 \]ewlineewlineWhere \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume. ewlineConsider this: if you pressurize a balloon, its volume decreases because the gas molecules are forced closer together. Conversely, if you reduce the pressure, the balloon inflates as the gas molecules spread out. ewlineUnderstanding this inverse relationship allows us to predict how a gas will react when we change its pressure.
Temperature Conversion
To use the Ideal Gas Law effectively, temperatures must be in Kelvin, not Celsius. The Kelvin scale starts at absolute zero, so it provides a more precise measure for scientific calculations. ewlineTo convert Celsius to Kelvin, use the formula: ewline\[ T(K) = T(^{\text{°}\text{C}}) + 273.15 \]ewlineewlineLet's look at the conversions from our exercise for better understanding:
  • For 200°C: \[ 200 + 273.15 = 473.15 \text{ K} \]
  • For 100°C: \[ 100 + 273.15 = 373.15 \text{ K} \]
  • For -73°C: \[ -73 + 273.15 = 200.15 \text{ K} \]
  • For 127°C: \[ 127 + 273.15 = 400.15 \text{ K} \]
These conversions are crucial because the Ideal Gas Law relies on Kelvin for its calculations to maintain consistency and accuracy in dealing with thermodynamic processes.
Gas Laws
Gas laws encompass various principles that describe how gases behave under different conditions of temperature, pressure, and volume. Besides the Ideal Gas Law, other specific laws provide more focused insights:
  • Boyle's Law: This law states that at constant temperature, the volume and pressure of a gas are inversely related: \[ P_1V_1 = P_2V_2 \]
  • Charles's Law: Charles's Law states that at constant pressure, the volume of a gas is directly proportional to its temperature: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
  • Avogadro's Law: It says that equal volumes of gases, at the same temperature and pressure, have the same number of molecules: \[ V \propto n \]
  • Combined Gas Law: This law combines Boyle's, Charles's, and Gay-Lussac's laws into one expression, relating pressure, volume, and temperature: \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \]
These laws are fundamental for understanding and predicting gas behavior in various physical and chemical processes.

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Most popular questions from this chapter

Popcorn pops because the horny endosperm, a tough, elastic material, resists gas pressure within the heated kernel until the pressure reaches explosive force. A 0.25 -mL kernel has a water content of \(1.6 \%\) by mass, and the water vapor reaches \(170^{\circ} \mathrm{C}\) and 9.0 atm before the kernel ruptures. Assume that water vapor can occupy \(75 \%\) of the kernel's volume. (a) What is the mass (in g) of the kernel? (b) How many milliliters would this amount of water vapor occupy at \(25^{\circ} \mathrm{C}\) and 1.00 atm?

Aqueous sulfurous acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{3}\right)\) was made by dissolving \(0.200 \mathrm{~L}\) of sulfur dioxide gas at \(19^{\circ} \mathrm{C}\) and \(745 \mathrm{mmHg}\) in water to yield \(500.0 \mathrm{~mL}\) of solution. The acid solution required \(10.0 \mathrm{~mL}\) of sodium hydroxide solution to reach the titration end point. What was the molarity of the sodium hydroxide solution?

What is the effect of the following on the volume of \(1 \mathrm{~mol}\) of an ideal gas? (a) The pressure changes from 760 torr to \(202 \mathrm{kPa}\), and the temperature changes from \(37^{\circ} \mathrm{C}\) to \(155 \mathrm{~K}\) (b) The temperature changes from \(305 \mathrm{~K}\) to \(32^{\circ} \mathrm{C},\) and the pressure changes from 2 atm to \(101 \mathrm{kPa}\). (c) The pressure is reduced by a factor of 4 (at constant \(T\) ).

A student states Boyle's law as follows: "The volume of a gas is inversely proportional to its pressure." How is this statement incomplete? Give a correct statement of Boyle's law.

What is the effect of the following on the volume of \(1 \mathrm{~mol}\) of an ideal gas? (a) The pressure is tripled (at constant \(T\) ). (b) The absolute temperature is increased by a factor of 3.0 (at constant \(P\) ). (c) Three more moles of the gas is added (at constant \(P\) and \(T\) ).

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