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Which will increase the density of a gas: doubling its temperature or doubling its pressure?

Short Answer

Expert verified
Answer: Doubling the pressure results in a greater increase in the density of a gas.

Step by step solution

01

Deriving a formula for density

To start, let's consider that the mass of the gas can be written as n times m, where m is the molar mass (n = mass/m). We can then substitute this expression for n into the ideal gas law equation: PV = (mass/m) RT => mPV = mass RT => mass/V = (mP)/(RT) Now we have an equation relating to density (mass/V): 蟻 = (mP)/(RT)
02

Analyzing the given changes

We are asked to consider two changes: doubling the temperature and doubling the pressure. Let's look at what happens to the density of the gas in both cases. Case 1: Doubling the temperature 蟻' = (mP)/(R(2T)) = (1/2)(mP)/(RT) This gives us half the original density: rho/2 Case 2: Doubling the pressure 蟻'' = (m(2P))/(RT) = (2mP)/(RT) This gives us double the original density: 2rho
03

Comparing the effects on density

Finally, let's compare the effects of both cases on the density of the gas: Case 1: Doubling the temperature results in halving the original density (蟻'/蟻 = 0.5). Case 2: Doubling the pressure results in doubling the original density (蟻''/蟻 = 2).
04

Answering the question

From the comparison of the effects of the two given changes, we can conclude that doubling the pressure (Case 2) results in an increase in gas density, whereas doubling the temperature (Case 1) results in a decrease in gas density. Therefore, doubling the pressure will increase the density of a gas.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a vital principle that helps us understand how gases behave under various conditions. The law is usually expressed by the equation \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. This equation gives us a comprehensive view of the relationship between a gas's pressure, volume, and temperature.
By manipulating this formula, it's possible to derive expressions that can help calculate different properties of gases, such as density. Understanding the ideal gas law allows us to predict how changes in pressure, temperature, or volume will affect the state of a gas. It's important to note that the ideal gas law assumes gases behave ideally, meaning intermolecular forces and the volume of the gas particles themselves are negligible. This is generally a good approximation under many conditions, but may not be entirely accurate at very high pressures or low temperatures.
Pressure Effect on Gas
Pressure plays a critical role in determining the behavior of a gas. According to the ideal gas law, pressure is directly proportional to the number of moles and temperature and inversely proportional to volume.
When the pressure of a gas increases, assuming constant temperature and moles, the volume typically decreases. This results in the gas particles being compressed into a smaller space, which in turn increases the density of the gas.
Doubling the pressure leads to a proportional increase in density, as seen in the formula derived from the ideal gas law: \( \rho = \frac{mP}{RT} \). If the pressure (\( P \)) is doubled, so is the density (\( \rho \)), assuming temperature and molar mass remain constant. This inverse relationship between pressure and volume, under constant temperature, is also classically illustrated by Boyle's Law.
Temperature Effect on Gas
Temperature is another crucial factor affecting gas behavior, guided by the ideal gas law. In this context, temperature must always be measured in Kelvin to ensure direct proportionality with the other variables.
As the temperature of a gas increases, and if pressure remains constant, the volume expands. This occurs because the gas particles gain kinetic energy with rising temperature, causing them to move more vigorously and spread apart from each other.
When examining the temperature's effect on density, doubling the temperature, while keeping pressure constant, results in the halving of gas density. This is evident from the density formula \( \rho = \frac{mP}{RT} \). With a doubled temperature (\(2T\)), the denominator increases, thereby reducing the overall density by half.
  • High temperature = Greater particle movement
  • More movement = Less density (assuming constant pressure)
  • Understanding this effect helps comprehend scenarios like how gases become less dense as they heat and rise, a principle applied in balloons and convection currents.

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Most popular questions from this chapter

Radon and helium are both by-products of the radioactive decay of uranium minerals. A fresh sample of carnotite, \(\mathrm{K}_{2}\left(\mathrm{UO}_{2}\right)_{2}\left(\mathrm{VO}_{4}\right)_{2}\) \(3 \mathrm{H}_{2} \mathrm{O},\) is put on display in a museum. Calculate the relative rates of diffusion of helium and radon under fixed conditions of pressure and temperature. Which gas diffuses more rapidly through the display case?

One cotton ball soaked in ammonia and another soaked in hydrochloric acid were placed at opposite ends of a \(1.00 \mathrm{m}\) glass tube (Figure \(\mathrm{P} 6.160\) ). The vapors diffused toward the middle of the tube and formed a white ring of ammonium chloride where they met. a. Write the chemical equation for this reaction. b. Will the ammonium chloride ring form closer to the end of the tube with ammonia or the end with hydrochloric acid? Explain your answer. c. Calculate the distance from the ammonia end to the position of the ammonium chloride ring.

Which of the following actions would produce the greater increase in the volume of a gas sample: (a) doubling the amount of gas in the sample at constant temperature and pressure or (b) raising the temperature from \(244^{\circ} \mathrm{C}\) to \(1100^{\circ} \mathrm{C}\) at constant pressure?

A typical blood pressure in a resting adult is "120 over 80," meaning 120 mmHg with each beat of the heart and \(80 \mathrm{mmHg}\) of pressure between heartbeats. Express these pressures in the following units: (a) torr; (b) \(\operatorname{atm} ;(\text { c })\) bar; \((\text { d) } \mathrm{kPa}\).

A sample of a gas enclosed in a cylinder with a piston has a volume of \(500.0 \mathrm{mL}\) at \(30.0^{\circ} \mathrm{C} .\) What is the sample volume at \(100.0^{\circ} \mathrm{C} ?\)

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