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A sample of nitrogen \(\left(\mathrm{N}_{2}\right)\) has a volume of \(50.0 \mathrm{~L}\) at a pressure of \(760 . \mathrm{mmHg}\). What is the volume, in liters, of the gas at each of the following pressures, if there is no change in temperature and amount of gas? a. \(1500 \mathrm{mmHg}\) b. \(4.00 \mathrm{~atm}\) c. \(0.500 \mathrm{~atm}\)

Short Answer

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a) 25.3 L; b) 12.5 L; c) 100 L

Step by step solution

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01

- Understand Boyle's Law

Boyle's Law states that the volume of a gas is inversely proportional to its pressure when the temperature and quantity of gas remain constant. This can be expressed as: \[ P_1 V_1 = P_2 V_2 \]
02

- Identify given values

The initial volume (\(V_1\)) is \(50.0\) L. The initial pressure (\(P_1\)) is \(760\) mmHg.
03

- Solve for new volume at \(1500 \,\text{mmHg}\)

Convert pressures into same unit if needed:\(P_2 = 1500\) mmHg.Using Boyle’s law \( P_1 V_1 = P_2 V_2\): \[ 760 \, mmHg \times 50.0 \, L = 1500 \, mmHg \times V_2 \]Solve for \(V_2\): \[ V_2 = \frac{760 \, mmHg \times 50.0 \, L}{1500 \, mmHg} \] \[ V_2 \approx 25.3 \, L \]
04

- Solve for new volume at \(4.00 \,\text{atm}\)

Convert \(4.00\) atm to mmHg since \(1\) atm = \(760\) mmHg. Therefore, \(P_2 = 4.00\) atm \(× 760\) mmHg/atm = \(3040\) mmHg.Using Boyle’s law \( P_1 V_1 = P_2 V_2 \): \[ 760 \, mmHg \times 50.0 \, L = 3040 \, mmHg \times V_2 \]Solve for \(V_2\): \[ V_2 = \frac{760 \, mmHg \times 50.0 \, L}{3040 \, mmHg} \] \[ V_2 \approx 12.5 \, L \]
05

- Solve for new volume at \(0.500 \,\text{atm}\)

Convert \(0.500\) atm to mmHg since \(1\) atm = \(760\) mmHg. Therefore, \(P_2 = 0.500\) atm \(× 760\) mmHg/atm = \(380\) mmHg.Using Boyle’s law \( P_1 V_1 = P_2 V_2 \): \[ 760 \, mmHg \times 50.0 \, L = 380 \, mmHg \times V_2 \]Solve for \(V_2\): \[ V_2 = \frac{760 \, mmHg \times 50.0 \, L}{380 \, mmHg} \] \[ V_2 = 100 \, L \]

Key Concepts

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

Gas Laws
Gas laws help us understand how gases behave under different conditions. Boyle's Law is one such principle. These laws are essential for many fields of science and engineering. Gas laws describe the relationships between volume, temperature, pressure, and quantity of a gas.
They provide the theoretical basis for understanding real-world phenomena. For instance, when you breathe, gas laws help explain how air flows in and out of your lungs.
Boyle's Law is particularly focused on the relationship between pressure and volume, simplifying the calculations when temperature and the amount of gas remain constant.
Pressure-Volume Relationship
Boyle's Law specifically addresses the pressure-volume relationship of a gas. It states that the volume of a gas is inversely proportional to its pressure. This means that if the pressure increases, the volume decreases, and vice versa. The formula is represented as: \[ P_1 V_1 = P_2 V_2 \] Where \(P_1\) is the initial pressure, \(V_1\) is the initial volume, \(P_2\) is the final pressure, and \(V_2\) is the final volume. This equation allows us to calculate the new volume when the gas pressure changes. Understanding this relationship helps in various applications such as in syringes, hydraulic systems, and understanding atmospheric behaviors. To solve any problem involving Boyle's Law, you just need two sets of known pressure and volume values.
Inverse Proportionality
Inverse proportionality is a core concept to grasp in Boyle’s Law. When two variables are inversely proportional, as one variable increases, the other decreases proportionally. For Boyle's Law, this means that if you double the pressure exerted on a gas, its volume will be halved given constant temperature and amount of gas. This relationship can be visualized through experiments and practical examples. For example, if you squeeze a balloon (increasing the pressure), the balloon's volume decreases. Understanding inverse proportionality helps to make predictions about how systems behave. In Boyle's Law, it lets us predict how a gas will respond to changes in pressure, making this concept very useful in daily life and scientific research.

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Most popular questions from this chapter

Solve for the new pressure, in torr, for each of the following, if \(n\) and \(V\) are constant: a. A gas with an initial pressure of 1200 torr at \(155^{\circ} \mathrm{C}\) is cooled to \(0{ }^{\circ} \mathrm{C}\). b. A gas in an aerosol can with an initial pressure of \(1.40\) atm at \(12^{\circ} \mathrm{C}\) is heated to \(35^{\circ} \mathrm{C}\).

A \(100.0\) -mL bubble of hot gases at \(225^{\circ} \mathrm{C}\) and \(1.80 \mathrm{~atm}\) is emitted from an active volcano. What is the new volume, in milliliters, of the bubble outside the volcano where the temperature is \(-25^{\circ} \mathrm{C}\) and the pressure is \(0.80 \mathrm{~atm}\), if the amount of gas remains the same?

Your spaceship has docked at a space station above Mars. The temperature inside the space station is a carefully controlled \(24{ }^{\circ} \mathrm{C}\) at a pressure of \(745 \mathrm{mmHg}\). A balloon with a volume of \(425 \mathrm{~mL}\) drifts into the airlock where the temperature is \(-95^{\circ} \mathrm{C}\) and the pressure is \(0.115 \mathrm{~atm}\). What is the new volume of the balloon \((n\) remains constant)? Assume that the balloon is very elastic.

Calculate the new temperature, in degrees Celsius, for each of the following with \(n\) and \(V\) constant: a. A sample of xenon at \(25^{\circ} \mathrm{C}\) and \(740 \mathrm{mmHg}\) is cooled to give a pressure of \(620 \mathrm{mmHg}\). b. A tank of argon gas with a pressure of \(0.950\) atm at \(-18{ }^{\circ} \mathrm{C}\) is heated to give a pressure of 1250 torr.

Use the kinetic molecular theory of gases to explain each of the following: a. Gases move faster at higher temperatures. b. Gases can be compressed much more than liquids or solids.

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