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Chapter 10: Problem 12

We have a jar \(A\) filled with gas characterized by parameters \(P, V\), and \(T\) and another jar \(B\) filled with gas with parameters \(2 P, V / 4\), and \(2 T\), where the symbols have their usual meanings. The ratio of the number of molecules of jar \(A\) to those of jar \(B\) is (A) \(1: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(4: 1\)

Short Answer

Expert verified
The ratio of the number of molecules of jar A to those of jar B is 2:1 (C).

Step by step solution

01

Ideal gas law for Jar A

We begin by writing the ideal gas law for Jar A: \(P_A V_A = n_A R T_A\).
02

Ideal gas law for Jar B

Next, we write the ideal gas law for Jar B: \(P_B V_B = n_B R T_B\).
03

Divide ideal gas laws

Divide the ideal gas law for Jar A by the ideal gas law for Jar B to find the ratio of their number of molecules: \[\frac{P_A V_A}{P_B V_B} = \frac{n_A R T_A}{n_B R T_B}\]
04

Cancel R

Since R is common in both the numerator and the denominator, cancel out the R terms in the equation: \[\frac{P_A V_A}{P_B V_B} = \frac{n_A T_A}{n_B T_B}\]
05

Substitute values

Substitute the given values for the parameters in Jar A and B: \[\frac{P_A V_A}{2P_A (V_A/4)} = \frac{n_A T_A}{n_B (2T_A)}\]
06

Simplify and find the ratio

Simplify the equation and find the ratio of the number of molecules in Jar A to those in Jar B: \[\frac{1}{2/4} = \frac{n_A}{n_B}\] \[\frac{1}{1/2} = \frac{n_A}{n_B}\] \[(2) = \frac{n_A}{n_B}\] So the ratio of the number of molecules of jar A to those of jar B is 2:1, which is answer choice (C) .

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Key Concepts

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

Ideal Gas Equation
The Ideal Gas Equation is a fundamental equation in chemistry and physics that relates the pressure (P), volume (V), temperature (T), and amount of substance (n) of an ideal gas. The equation is expressed as:
\[ PV = nRT \]
where R is the universal gas constant. This equation assumes that the gas particles are point particles that do not interact with each other except during elastic collisions. In real-world applications, this equation is an excellent approximation for most gases under normal conditions.
When solving ideal gas law problems, it's essential to ensure that the units used for pressure, volume, and temperature are compatible with the units for R. In the exercise provided, the ideal gas law helps us understand how changing conditions in two different jars affect the number of molecules, given all gases behave ideally.
Molar Volume
Molar volume is the volume occupied by one mole of a substance, usually a gas, at a given temperature and pressure. Under standard temperature and pressure conditions (0 degrees Celsius and 1 atmosphere), the molar volume of an ideal gas is approximately 22.4 liters per mole. This concept is essential because it allows us to relate the volume of a gas to the amount of substance in moles when dealing with ideal gases.
In the context of the exercise, we use molar volume indirectly when comparing the volume and amount of gas in the two jars, even though the actual value of the molar volume isn't needed in this calculation. The proportionality provided by the ideal gas law allows us to compare the volume and mole ratios under varying conditions without explicitly calculating molar volumes.
Avogadro's Law
Avogadro's Law states that equal volumes of ideal or perfect gases, at the same temperature and pressure, contain the same number of molecules. It is mathematically stated as:
\[ V \propto n \]
(where V is the volume and n is the amount in moles), assuming that temperature and pressure remain constant. This law forms the basis for the molar volume of a gas. Avogadro's Law is a specific case of the ideal gas law when dealing with volumes and quantities in moles.
In our exercise, Avogadro's Law underpins the calculations. By understanding that the relationship between the number of gas molecules and volume is direct, we can deductively solve for the ratio of the number of molecules between the two jars. As the exercise demonstrates, this relationship is crucial when comparing quantities of gas in different conditions.

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Most popular questions from this chapter

1 mole of an ideal gas is contained in a cubical volume \(V\), ABCDEFGH at \(300 \mathrm{~K}\) (Fig. 10.18). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time, (A) The pressure on \(E F G H\) would be zero. (B) The pressure on all the faces will be equal. (C) The pressure of \(E F G H\) would be double the pressure on \(A B C D\). (D) The pressure of \(E F G H\) would be half that on \(A B C D .\)

A bimetallic strip is formed by two identical strips, one of copper and the other of brass. The coefficients of linear expansion of the two metals are \(\alpha_{C}\) and \(\alpha_{B}\). On heating, the temperature of the strip goes up by \(\Delta T\) and the strip bends to form an arc of radius of curvature \(R\). Then \(R\) is (A) proportional to \(\Delta T\). (B) inversely proportional to \(\Delta T\). (C) proportional to \(\left|\alpha_{B}-\alpha_{C}\right|\). (D) inversely proportional to \(\left|\alpha_{B}-\alpha_{C}\right|\).

Heat given to a body which raises its temperature by \(1^{\circ} \mathrm{C}\) is (A) water equivalent. (B) thermal capacity. (C) specific heat. (D) temperature gradient.

If \(S\) is stress and \(Y\) is Young's modulus of material of a wire, then the energy stored in the wire per unit volume is (A) \(2 S^{2} Y\) (B) \(S^{2} / 2 Y\) (C) \(2 Y / S^{2}\) (D) \(S / 2 Y\)

A non-conducting piston of mass \(m\) and area \(S_{0}\) divides a non-conducting, closed cylinder as shown in Fig. \(10.22\). Piston having mass \(m\) is connected with top wall of cylinder by a spring of force constant \(k\). Top part is evacuated and bottom part contains an ideal gas at pressure \(P_{0}\) in equilibrium position. Adiabatic constant \(\gamma\) and in equilibrium length of each part is \(l\). (neglect friction) Find angular frequency for small oscillation. (A) \(\sqrt{\frac{k l+\gamma P_{0} S_{0}}{2 m l}}\) (B) \(\sqrt{\frac{2 k l+\gamma P_{0} S_{0}}{m l}}\) (C) \(\sqrt{\frac{k}{m}}\) (D) \(\sqrt{\frac{k l+\gamma P_{0} S_{0}}{m l}}\)
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