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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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Most popular questions from this chapter

The radius of a metal sphere at room temperature \(T\) is \(R\) and the coefficient of linear expansion of the metal is \(\alpha .\) The sphere heated a little by a temperature \(\Delta T\) so that its new temperature is \(T+\Delta T\). The increase in the volume of the sphere is approximately. (A) \(2 \pi R \alpha \Delta T\) (B) \(\pi R^{2} \alpha \Delta T\) (C) \(4 \pi R^{3} \alpha \Delta T / 3\) (D) \(4 \pi R^{3} \alpha \Delta T\)

A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly, (A) its speed of rotation increases. (B) its speed of rotation decreases. (C) its speed of rotation remains same. (D) its speed in increases because its moment of inertia increases.

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