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The kinetic energy of 4 moles of nitrogen gas at \(127^{\circ} \mathrm{C}\) is ....... Kcals. \(\left(\mathrm{R}=2 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) 4400 (b) 3200 (c) 4800 (d) 1524

Short Answer

Expert verified
The kinetic energy is 4.8 kcal, which matches option (c) 4800 cal.

Step by step solution

01

Convert Temperature to Kelvin

First, we need to convert the given temperature from Celsius to Kelvin using the formula: \[ T(K) = T(\degree C) + 273 \]Given temperature is \(127^{\circ}\)C, so the conversion is:\[ T = 127 + 273 = 400 \text{ K} \]
02

Apply the Kinetic Energy Formula

The average kinetic energy per mole of an ideal gas is given by:\[ KE_{\text{per mole}} = \frac{3}{2}RT \]where:- \( R \) is the universal gas constant \( = 2 \text{ cal} \text{ mol}^{-1} \text{ K}^{-1} \)- \( T \) is the temperature in Kelvin \( = 400 \text{ K} \)
03

Calculate Kinetic Energy Per Mole

Substitute the values into the kinetic energy formula:\[ KE_{\text{per mole}} = \frac{3}{2} \times 2 \text{ cal} \cdot 400 \]\[ KE_{\text{per mole}} = 3 \times 400 = 1200 \text{ cal/mol} \]
04

Determine Total Kinetic Energy for 4 Moles

To find the total kinetic energy for 4 moles of the gas, multiply the kinetic energy per mole by the number of moles:\[ KE_{\text{total}} = 1200 \text{ cal/mol} \times 4 \text{ moles} \]\[ KE_{\text{total}} = 4800 \text{ cal} \]
05

Convert calories to Kilocalories

Since 1 kilocalorie equals 1000 calories, convert the total energy:\[ 4800 \text{ cal} = \frac{4800}{1000} \text{ kcal} = 4.8 \text{ kcal} \]

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental principle in chemistry and physics that describes the behavior of an ideal gas. The law combines several individual gas laws (Boyle's, Charles's, and Avogadro's laws) to form the equation:\[ PV = nRT \]where:
  • P is the pressure of the gas
  • V is the volume
  • n is the number of moles
  • R is the universal gas constant, which can differ in units such as liter atmospheres or calories
  • T is the temperature in Kelvin
This equation allows us to relate the physical properties of the gas. In many problems, you need to manipulate this equation to find one value when the others are known. The Ideal Gas Law assumes a hypothetical gas where the molecules have no volume and do not interact. It's essential to ensure temperature is in Kelvin for correctness, as this scale starts at absolute zero where gases theoretically have no motion.
Temperature Conversion
Understanding how to convert temperature between different units is crucial when working with gases, especially in chemistry. The most common conversion is between Celsius and Kelvin.To convert Celsius to Kelvin, use the formula:\[ T(K) = T(^{\circ}C) + 273 \]In our exercise, a given temperature of \(127^{\circ}C\) converts as follows:\[ T = 127 + 273 = 400 \text{ K} \]Using Kelvin is vital for calculations involving the Ideal Gas Law because this scale is directly proportional to the kinetic energy of particles. Unlike Celsius, the Kelvin scale doesn't have negative numbers which ensures calculations remain mathematically consistent, especially when applied to the principles of kinetic energy and gas laws.
Energy Units Conversion
Energy can be expressed in different units, and converting between them can be necessary in various calculations. In the context of kinetic energy for ideal gases, we often convert between calories and kilocalories.1 calorie is the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius. Meanwhile, 1 kilocalorie (often referred to as a Calorie in food energy) is equal to 1000 calories.To convert calories to kilocalories, use:\[ \text{kcal} = \frac{\text{cal}}{1000} \]Taking our exercise as an example, 4800 calories is converted to:\[ 4800 \text{ cal} = \frac{4800}{1000} \text{ kcal} = 4.8 \text{ kcal} \]This conversion is particularly helpful in energy calculations related to thermodynamics and physical chemistry where large energy values are common. By using kilocalories, values become more manageable and easier to interpret.

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Most popular questions from this chapter

Which of the following mixture of gases does not obey Dalton's law of partial pressure? (a) \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) (b) \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) (c) \(\mathrm{Cl}_{2}\) and \(\mathrm{SO}_{2}\) (d) \(\mathrm{NH}_{3}\) and \(\mathrm{HCl}\)

If \(\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3} \ldots \ldots \ldots\) represents the speed of \(\mathrm{n}_{1}\), \(\mathrm{n}_{2}, \mathrm{n}_{3}, \ldots\) molecules, then the root mean square of speed is (a) \(\left(\frac{\mathrm{n}_{1} \mathrm{C}_{1}^{2}+\mathrm{n}_{2} \mathrm{C}_{2}^{2}+\mathrm{n}_{3} \mathrm{C}_{3}^{2}+\ldots}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}+\ldots}\right)^{1 / 2}\) (b) \(\left(\frac{n_{1} C_{1}^{2}+n_{2} C_{2}^{2}+n_{3} C_{3}^{2}+\ldots}{n_{1}+n_{2}+n_{3}+\ldots}\right)^{2}\) (c) \(\frac{\left(\mathrm{n}_{1} \mathrm{C}_{1}^{2}\right)^{1 / 2}}{\mathrm{n}_{1}}+\frac{\left(\mathrm{n}_{2} \mathrm{C}_{2}^{2}\right)^{1 / 2}}{\mathrm{n}_{2}}+\frac{\left(\mathrm{n}_{3} \mathrm{C}_{3}^{2}\right)^{1 / 2}}{\mathrm{n}_{3}}+\ldots\) (d) \(\left[\frac{\left(\mathrm{n}_{1} \mathrm{C}_{1}+\mathrm{n}_{2} \mathrm{C}_{2}+\mathrm{n}_{3} \mathrm{C}_{3}+\ldots\right)^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}+\ldots}\right]^{1 / 2}\)

Equal weights of ethane and hydrogen are mixed in an empty container at \(25^{\circ} \mathrm{C}\). The fraction to total pressure exerted by hydrogen is (a) \(1: 2\) (b) \(1: 1\) (c) \(1: 16\) (d) \(15: 16\)

At a certain pressure, volume of a gas at \(27^{\circ} \mathrm{C}\) is 20 litre. If the pressure and temperature are doubled, its volume will be (a) 20 litre (b) 40 litre (c) \(8.2\) litre (d) \(10.9\) litre

The values of van der Waals constant ' \(\alpha\) ' for the gases \(\mathrm{O}_{2}, \mathrm{~N}_{2}, \mathrm{NH}_{3}\) and \(\mathrm{CH}_{4}\) are \(1.360,1.390,4.170\) and \(2.253\) L atm. mol \(^{2}\) respectively. The gas which can most easily be liquefied is (a) \(\mathrm{O}_{2}\) (b) \(\mathrm{N}_{2}\) (c) \(\mathrm{NH}_{3}\) (d) \(\mathrm{CH}_{4}\)

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