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Chapter 19: Problem 48

At room temperature, identical gas cylinders contain 10 moles of nitrogen gas and argon gas, respectively. Determine the ratio of energies stored in the two systems. Assume ideal gas behavior.

Short Answer

Expert verified
Answer: The ratio of energies stored in the Nitrogen and Argon gas cylinders is 5/3 (approximately 1.67).

Step by step solution

01

Recall the Ideal Gas Internal Energy formula

To find the energy stored in an ideal gas, we can use the formula for Internal Energy (U) of an ideal gas system: U = n * C_v * T where n is the number of moles, C_v is the molar specific heat at constant volume, and T is the temperature in Kelvin.
02

Define the given values

We are given the following information: - Each cylinder contains 10 moles of gas (n = 10 moles). - Nitrogen (N2) and Argon (Ar) gases are diatomic and monatomic, respectively. - Room temperature is assumed to be 298 K.
03

Find the molar specific heat for Nitrogen and Argon gases

For a diatomic ideal gas (such as Nitrogen, N2), the molar specific heat at constant volume (C_v) is given by: C_v_diatomic = \(\frac{5}{2}R\) For a monatomic ideal gas (such as Argon, Ar), the molar specific heat at constant volume (C_v) is given by: C_v_monatomic = \(\frac{3}{2}R\) Where R is the Universal gas constant, approximately 8.314 J/mol*K.
04

Calculate the internal energy of Nitrogen and Argon gases

Now, we can find the internal energy stored in each cylinder. Using the formula U = n * C_v * T and substituting the given values, we get: For Nitrogen (N2): U_N2 = n * C_v_diatomic * T = 10 * \(\frac{5}{2}R\) * 298 For Argon (Ar): U_Ar = n * C_v_monatomic * T = 10 * \(\frac{3}{2}R\) * 298 Now simplify: U_N2 = 10 * \(\frac{5}{2}(8.314)\) * 298 U_Ar = 10 * \(\frac{3}{2}(8.314)\) * 298
05

Find the energy ratio

Finally, to find the ratio of energies stored in the Nitrogen and Argon gas cylinders, divide U_N2 by U_Ar: Energy Ratio = \(\frac{U_{N_{2}}}{U_{Ar}}\) = \(\frac{10 * \frac{5}{2}(8.314) * 298}{10 * \frac{3}{2}(8.314) * 298}\) Since the factors 10, 8.314, and 298 appear in both numerator and denominator, we can cancel them out: Energy Ratio = \(\frac{\frac{5}{2}}{\frac{3}{2}}\) Now simplify: Energy Ratio = \(\frac{5}{3}\) So, the ratio of energies stored in Nitrogen and Argon gas cylinders is \(\frac{5}{3}\) (approximately 1.67).

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Most popular questions from this chapter

As noted in the text, the speed distribution of molecules in the Earth's atmosphere has a significant impact on its composition. a) What is the average speed of a nitrogen molecule in the atmosphere, at a temperature of \(18.0^{\circ} \mathrm{C}\) and a (partial) pressure of \(78.8 \mathrm{kPa} ?\) b) What is the average speed of a hydrogen molecule at the same temperature and pressure?

In a period of \(6.00 \mathrm{~s}, 9.00 \cdot 10^{23}\) nitrogen molecules strike a section of a wall with an area of \(2.00 \mathrm{~cm}^{2}\). If the molecules move with a speed of \(400.0 \mathrm{~m} / \mathrm{s}\) and strike the wall head on in elastic collisions, what is the pressure exerted on the wall? (The mass of one \(\mathrm{N}_{2}\) molecule is \(4.68 \cdot 10^{-26} \mathrm{~kg}\).)

An ideal gas has a density of \(0.0899 \mathrm{~g} / \mathrm{L}\) at \(20.00^{\circ} \mathrm{C}\) and \(101.325 \mathrm{kPa}\). Identify the gas.

What is the approximate energy required to raise the temperature of \(1.00 \mathrm{~L}\) of air by \(100 .{ }^{\circ} \mathrm{C} ?\) The volume is held constant.

Liquid nitrogen, which is used in many physics research labs, can present a safety hazard if a large quantity evaporates in a confined space. The resulting nitrogen gas reduces the oxygen concentration, creating the risk of asphyxiation. Suppose \(1.00 \mathrm{~L}\) of liquid nitrogen \(\left(\rho=808 \mathrm{~kg} / \mathrm{m}^{3}\right)\) evaporates and comes into equilibrium with the air at \(21.0^{\circ} \mathrm{C}\) and \(101 \mathrm{kPa}\). How much volume will it occupy?
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