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What is the effect of the following on the volume of \(1 \mathrm{~mol}\) of an ideal gas? (a) Temperature decreases from \(800 \mathrm{~K}\) to \(400 \mathrm{~K}\) (at constant \(P\) ). (b) Temperature increases from \(250^{\circ} \mathrm{C}\) to \(500^{\circ} \mathrm{C}\) (at constant \(P\) ). (c) Pressure increases from 2 atm to 6 atm (at constant \(T\) ).

Short Answer

Expert verified
a) Volume is halved. b) Volume increases by about 1.48 times. c) Volume is reduced to one-third.

Step by step solution

01

- Understand the Ideal Gas Law

The Ideal Gas Law is given by: \[ PV = nRT \]where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
02

- Analyze Part (a)

For part (a), the temperature decreases from 800 K to 400 K at constant pressure (P). According to the Ideal Gas Law:\[ V = \frac{nRT}{P} \]Since \( n \), \( R \), and \( P \) are constant, the volume is directly proportional to the temperature. Hence:\[ V_1 / T_1 = V_2 / T_2 \]\[ V_2 = V_1 \times \frac{T_2}{T_1} = V_1 \times \frac{400}{800} = \frac{1}{2} V_1 \]This indicates that the volume is halved.
03

- Analyze Part (b)

For part (b), the temperature increases from 250 °C to 500 °C at constant pressure. Convert temperatures to Kelvin (K) first:\[ 250^{\circ} \mathrm{C} = 523 \mathrm{K} \] \[ 500^{\circ} \mathrm{C} = 773 \mathrm{K} \] Using the same direct proportionality principle:\[ V_2 = V_1 \times \frac{T_2}{T_1} = V_1 \times \frac{773}{523} \approx 1.48 V_1 \]The volume increases by a factor of approximately 1.48.
04

- Analyze Part (c)

For part (c), the pressure increases from 2 atm to 6 atm at constant temperature. According to the Ideal Gas Law:\[ V = \frac{nRT}{P} \]The volume is inversely proportional to the pressure, hence:\[ V_1 \times P_1 = V_2 \times P_2 \]\[ V_2 = V_1 \times \frac{P_1}{P_2} = V_1 \times \frac{2}{6} = \frac{1}{3} V_1 \]This indicates that the volume is reduced to one-third.

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Key Concepts

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

Temperature Effect on Gas Volume
The volume of an ideal gas is directly proportional to its temperature when pressure is held constant.
According to the Ideal Gas Law, when temperature decreases, the volume also decreases, and vice versa.
This is expressed mathematically as: \[ V = nRT / P \]
If the temperature is halved, the volume is halved as well.
By examining our example, a temperature drop from 800 K to 400 K results in the volume being halved.
Conversely, an increase from 250°C (523 K) to 500°C (773 K) increases the volume by a factor of 1.48, calculated as: \[ V_2 = V_1 \times \frac{T_2}{T_1} \approx 1.48 V_1. \] Ensuring we convert temperatures to Kelvin before use ensures accuracy.
Temperature changes have a straightforward, linear effect on gas volume when pressure remains unchanged.
Pressure Effect on Gas Volume
Pressure significantly impacts gas volume, but in an inverse manner.
Using the Ideal Gas Law, when temperature and the number of moles are constant, volume and pressure have an inverse relationship.
This means if the pressure is increased, the volume decreases, and if the pressure is decreased, the volume increases.
The relationship is given by: \[ V = \frac{nRT}{P} \] For example, if the pressure is tripled from 2 atm to 6 atm, the volume will reduce to one-third of its initial value: \[ V_2 = V_1 \times \frac{P_1}{P_2} = V_1 \times \frac{2}{6} = \frac{1}{3}V_1. \]
Understanding this inverse proportionality helps predict how a gas will behave under varying pressures.
Proportionality in Ideal Gas Law
The Ideal Gas Law, given by \[ PV = nRT \] describes the relationship between pressure (P), volume (V), number of moles (n), ideal gas constant (R), and temperature (T).
It highlights the proportionality among these variables.
Volume is directly proportional to temperature and inversely proportional to pressure.
Keep in mind:
  • When temperature increases at constant pressure, volume increases.
  • When pressure increases at constant temperature, volume decreases.
Both scenarios are clear cases of direct and inverse proportionality under controlled conditions.
Being familiar with these proportionality rules is crucial for predicting and understanding the behavior of ideal gases in different conditions.

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Most popular questions from this chapter

A gaseous organic compound containing only carbon, hydrogen, and nitrogen is burned in oxygen gas, and the volume of each reactant and product is measured under the same conditions of temperature and pressure. Reaction of four volumes of the compound produces 4 volumes of \(\mathrm{CO}_{2}, 2\) volumes of \(\mathrm{N}_{2},\) and 10 volumes of water vapor. (a) How many volumes of \(\mathrm{O}_{2}\) were required? (b) What is the empirical formula of the compound?

A sample of an unknown gas effuses in 11.1 min. An equal volume of \(\mathrm{H}_{2}\) in the same apparatus under the same conditions effuses in \(2.42 \mathrm{~min} .\) What is the molar mass of the unknown gas?

An equimolar mixture of Ne and Xe is accidentally placed in a container that has a tiny leak. After a short while, a very small proportion of the mixture has escaped. What is the mole fraction of Ne in the effusing gas?

The gravitational force exerted by an object is given by \(F=m g\) where \(F\) is the force in newtons, \(m\) is the mass in kilograms, and \(g\) is the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) Use the definition of the pascal to calculate the mass (in \(\mathrm{kg}\) ) of the atmosphere above \(1 \mathrm{~m}^{2}\) of ocean. (b) Osmium \((Z=76)\) is a transition metal in Group \(8 \mathrm{~B}(8)\) and has the highest density of any element ( \(22.6 \mathrm{~g} / \mathrm{mL}\) ). If an osmium column is \(1 \mathrm{~m}^{2}\) in area, how high must it be for its pressure to equal atmospheric pressure? [Use the answer from part (a) in your calculation.]

The thermal decomposition of ethylene occurs during the compound's transit in pipelines and during the formation of polyethylene. The decomposition reaction is $$\mathrm{CH}_{2}=\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{C}(\mathrm{graphite})$$ If the decomposition begins at \(10^{\circ} \mathrm{C}\) and 50.0 atm with a gas density of \(0.215 \mathrm{~g} / \mathrm{mL}\) and the temperature increases by \(950 \mathrm{~K}\), (a) What is the final pressure of the confined gas (ignore the volume of graphite and use the van der Waals equation)? (b) How does the \(P V / R T\) value of \(\mathrm{CH}_{4}\) compare to that in Figure \(5.23 ?\) Explain.

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