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A closed container contains equal number of moles of two gases \(X\) and \(Y\) at a total pressure of \(710 \mathrm{~mm}\) of \(\mathrm{Hg}\). If gas \(X\) is removed from the mixture, the pressure will (a) become double (b) become half (c) remain same (d) become one-fourth.

Short Answer

Expert verified
The pressure will become half when gas X is removed from the mixture.

Step by step solution

01

Identify the Concept

The pressure of a gas mixture is directly proportional to the number of moles of the gas present, according to Dalton's Law of Partial Pressures. If the number of moles is unchanged, the pressure remains the same, and if the number of moles changes, the pressure changes proportionally.
02

Determine the Effect of Removal of Gas X

In the given problem, gas X is removed from a container that has equal moles of gas X and gas Y, each contributing equally to the total pressure. Removing gas X would eliminate its partial pressure contribution.
03

Calculate the New Pressure

Since the container initially contains equal moles of gases X and Y, the pressure contributed by gas X alone would be half of the total pressure. Removing gas X will therefore leave only gas Y, effectively halving the initial total pressure.

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Key Concepts

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

Chemistry and Dalton's Law
Understanding the behavior of gases in Chemistry involves various laws that describe how gases interact and behave under certain conditions. One fundamental principle is Dalton's Law of Partial Pressures, which states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

Imagine a closed container filled with a group of friends, each taking up their own space. If some friends leave, there's more room for the others to spread out. In our gaseous analogy, however, gases always fill the available space completely, so when one gas leaves the container, the remaining gas takes up the full space but with reduced pressure because there are fewer 'friends' or molecules exerting force.

The step-by-step solution to the textbook problem illustrates this concept. With the removal of gas X, only gas Y continues to exert pressure. Since gas X and Y were initially contributing equally, upon removal of X, gas Y alone contributes to the pressure, which is now half of what it was—showing a directly proportional relationship between the amount of gas and the pressure it exerts in a closed system.
Gases and Pressure Dynamics
When we talk about gases and pressure, it's important to understand that gases are composed of molecules that are in constant motion, colliding with the walls of their container. This is what creates pressure. It's like a constant buzz of activity at a party where guests are walking around and occasionally bumping into furniture (the walls of the container).

The greater the number of gas molecules (guests) in the container (party), the more collisions occur with the walls, and thus, the higher the pressure. If you reduce the number of molecules, like when we remove gas X in our problem, then you have fewer collisions and therefore less pressure.

This dynamic nature of gas molecules is why the pressure inside a container changes in response to the number of molecules and is a fundamental concept for understanding how gases behave.
Molar Concepts in Chemistry
Moles are a foundational concept in Chemistry, especially when dealing with gases. A mole is like a specific package size, sort of like buying rice in a 1 kg packet, that always contains the same number of entities—in this case, Avogadro's number, which is approximately 6.022 x 10^23. This allows chemists to count particles like atoms and molecules in a practical way.

When we relate moles to gases, each mole of an ideal gas at standard temperature and pressure occupies the same volume. This is why, in the textbook problem, we could reason that since there's an equal number of moles of gas X and Y in the container, they occupy equal volumes and exert equal pressure.

Understanding molar concepts helps to predict how the removal of a certain number of moles of a gas, such as gas X, will affect the overall pressure in a container—a key point in solving problems like the one presented in the textbook.

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Most popular questions from this chapter

In van der Waals equation for a non-ideal gas, the term that accounts for intermolecular force is (a) \((V-b)\) (b) \(\left(P+\frac{a}{V_{-}^{2}}\right)\) (c) \(R T\) (d) \(P V\)

At what temperature \(28 \mathrm{~g}\) of \(\mathrm{N}_{2}\) will occupy a volume of 20 litres at 2 atm? (a) \(300.0 \mathrm{~K}\) (b) \(487.2 \mathrm{~K}\) (c) \(289.6 \mathrm{~K}\) (d) \(283.8 \mathrm{~K}\)

The correct value of the gas constant \({ }^{\prime} R\) is close to (a) \(0.082\) litre-atmosphere \(\mathrm{K}\) (b) \(0.082\) litre-atmosphere \(\mathrm{K}^{-1} \mathrm{~mol}^{-1}\) (c) \(0.082\) litre-atmosphere \(^{-1} \mathrm{~K} \mathrm{~mol}^{-1}\) (d) \(0.082\) litre \(^{-1}\) atmosphere \(^{-1} \mathrm{~K} \mathrm{~mol}\).

Pick out the wrong statement(s). (i) Vapour pressure of a liquid is the measure of the strength of intermolecular attractive forces. (ii) Surface tension of a liquid acts perpendicular to the surface of the liquid. (iii) Vapour pressure of all liquids is same at their freezing points. (iv) Liquids with stronger intermolecular attractive forces are more viscous than those with weaker intermolecular force. (a) (ii), (iii) and (iv) (b) (ii) and (iii) (c) (i), (ii) and (iii) (d) (iii) only

Read the following statements and identify the incorrect statement. (a) Volume of one mole of a gas at critical temperature is called molar volume. (b) Pressure of a gas at critical temperature is called critical pressure. (c) The critical temperature, pressure and volume are called critical constants. (d) Critical temperature is the highest temperature at which a gas can exist as liquid, above this temperature it is a gas.

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