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How many times the average speed of the molecules in a gas becomes when the temperature is raised from \(27^{\circ} \mathrm{C}\) to \(159^{\circ} \mathrm{C}\) ? (a) \(1.2\) (b) \(1.44\) (c) \(5.89\) (d) \(2.43\)

Short Answer

Expert verified
The average speed of the molecules in the gas becomes \(1.2\times\) when the temperature is raised from \(27^\circ \mathrm{C}\) to \(159^\circ \mathrm{C}\).

Step by step solution

01

Understand the Relationship Between Speed and Temperature

The average speed of the molecules in a gas is directly proportional to the square root of its absolute temperature. This is described by the equation \( v \propto \sqrt{T} \), where \( v \) is the speed and \( T \) is the temperature in kelvins.
02

Convert Temperatures to Kelvin

To apply the equation, first convert the given temperatures from Celsius to Kelvin. Use the formula \( T(\mathrm{K}) = T(\mathrm{^\circ C}) + 273.15 \). For the initial temperature, \( T_i = 27^\circ \mathrm{C} = 27 + 273.15 = 300.15 \mathrm{K} \). For the final temperature, \( T_f = 159^\circ \mathrm{C} = 159 + 273.15 = 432.15 \mathrm{K} \).
03

Calculate the Ratio of the Speeds

Calculate the ratio of the final speed to the initial speed which is equal to the square root of the ratio of the final to initial temperatures in kelvin. \( \frac{v_f}{v_i} = \sqrt{\frac{T_f}{T_i}} \). Substitute the temperatures to find the ratio \( \frac{v_f}{v_i} = \sqrt{\frac{432.15}{300.15}} \) and calculate the value.
04

Calculate the Result

After substituting the values, the ratio is calculated as follows: \( \frac{v_f}{v_i} = \sqrt{\frac{432.15}{300.15}} \approx \sqrt{1.44} = 1.2 \).

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Key Concepts

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

Gas Laws
Understanding the behavior of gases under various conditions is critical to grasping foundational concepts in chemistry and physics. Gas laws describe these behaviors quantitatively and can predict how a gas will change state with variations in temperature, volume, and pressure.

The gas laws, including Boyle's Law, Charles's Law, and Gay-Lussac's Law, each highlight different aspects of these changes. For instance, Charles's Law states that the volume of a gas is directly proportional to its temperature when pressure is held constant. This fundamental principle is intrinsic to solving problems related to molecular speed as it underpins why increased temperatures can increase speed.

In terms of molecular speed, a rise in temperature as per Charles's Law means more energetic gas particles. Since the particles are moving faster, they collide more frequently and with greater force. These insights from gas laws help in solving complex problems by providing a clear framework within which the behavior of gases can be understood.
Kinetic Molecular Theory
The Kinetic Molecular Theory (KMT) provides a conceptual model for understanding the physical properties of gases. It explains that gases are composed of particles in constant random motion and that the behavior of these particles can help us understand properties like temperature, pressure, and volume.

According to KMT, temperature is a measure of the average kinetic energy of the gas particles. Thus, when temperature increases, the average kinetic energy of the particles also increases, leading to a higher velocity or speed of the gas molecules. This principle was employed in the provided exercise, revealing how the molecular speed of a gas correlates with changes in temperature.

In addition to the contribution to molecular speed, KMT is also fundamental in explaining phenomena such as diffusion, effusion, and the pressure exerted by gases. These concepts can be difficult, but by tying them to KMT, they become far more digestible for students.
Conversion of Temperature Units
In scientific problems, especially those that apply the gas laws or KMT, it is often necessary to convert temperature measurements from one unit to another. The Celsius and Kelvin scales are the most commonly used temperature scales in scientific calculations.

To convert Celsius to Kelvin, which is a crucial step in many gas law problems, you add 273.15 to the Celsius temperature. This shift ensures that we measure temperature from an absolute zero point, where theoretically, there is no particle movement.

During the solution of the exercise, temperature conversion allowed us to use Kelvin which is required for the equations of the gas laws to hold true. This step is foundational and cannot be overlooked, as using incorrect temperature units would result in flawed calculations. For students, mastering the art of converting temperature units is vital and could be the key to solving thermal dynamics problems accurately.

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Most popular questions from this chapter

A balloon is filled with \(\mathrm{N}_{2} \mathrm{O}\) is pricked with a sharp point and quickly plunged into a tank of \(\mathrm{CO}_{2}\) under the same pressure and temperature. The balloon will (a) be enlarged (b) shrink (c) remain unchanged in size (d) collapse completely

Pressure of \(1 \mathrm{~g}\) of an ideal gas \(\mathrm{A}\) at \(27^{\circ} \mathrm{C}\) is found to be 2 bar. When \(2 \mathrm{~g}\) of another ideal gas \(\mathrm{B}\) is introduced in the same flask at same temperature the pressure becomes 3 bar. What is the relationship between their molecular masses? (a) \(2 M_{\mathrm{B}}=M_{\mathrm{A}}\) (b) \(M_{\mathrm{B}}=M_{\mathrm{A}}\) (c) \(M_{\mathrm{B}}=4 M_{\mathrm{A}}\) (d) \(M_{\mathrm{B}}=2 M_{\mathrm{A}}\)

Vapour is injected at a uniform rate in a closed vessel which was initially evacuated. The pressure in the vessel (a) increases continuously (b) decreases continuously (c) first increases and then decreases (d) first increases and then becomes constant

The \(P V-P\) isotherms of 1 mole of different gases at \(273 \mathrm{~K}\), if the limit of pressure tending to zero, converge to a value of \(P V=\) (a) \(11.21\) -atm (b) \(22.41\) -atm (c) zero (d) \(22.41\)

At moderate pressure, the compressibility factor for a gas is given as: \(Z=1+0.35 P\) \(-\frac{168}{T} \cdot P\), where \(P\) is in bar and \(T\) is in Kelvin. What is the Boyle's temperature of the gas? (a) \(168 \mathrm{~K}\) (b) \(480 \mathrm{~K}\) (c) \(58.8 \mathrm{~K}\) (d) \(575 \mathrm{~K}\)

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