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Chapter 8: Problem 1125

One \(\mathrm{kg}\) of adiatomic gas is at a pressure of $5 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$ The density of the gas is \(\left\\{(5 \mathrm{~kg}) / \mathrm{m}^{3}\right\\}\) what is the energy of the gas due to its thermal motion ? (A) \(2.5 \times 10^{5} \mathrm{~J}\) (B) \(3.5 \times 10^{5} \mathrm{~J}\) (C) \(4.5 \times 10^{5} \mathrm{~J}\) (D) \(1.5 \times 10^{5} \mathrm{~J}\)

Short Answer

Expert verified
The short answer for this question is that the energy of the gas due to its thermal motion is \(1.5 \times 10^{5}\, J\). The correct choice is option (D).

Step by step solution

01

Find the Volume

To find the volume of the gas, we can use the definition of density: \(Density = \frac{Mass}{Volume}\). We are given the mass (\(M = 1 kg\)) and the density (\( \rho = 5 kg/m^3\)). Solving for volume we get \[V = \frac{M}{\rho} = \frac{1 kg}{5 kg/m^3} = 0.2 m^3.\]
02

Calculate the Thermal Energy

Now, we use the formula for the energy of a gas due to thermal motion: \(E_{th} = \frac{3}{2}P_{V}\). Here, \(P\) is the pressure and \(V\) is the volume. Plugging in the values \[E_{th} = \frac{3}{2} \times 5 \times 10^{5}\,N/m^2 \times 0.2\, m^3 = 1.5 \times 10^{5} J.\] So the energy of the gas due to its thermal motion is \(1.5 \times 10^{5} J\). The correct answer is option (D) \(1.5 \times 10^{5}\, J\).

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Most popular questions from this chapter

Work done per mole in an isothermal ? change is (A) RT log \(_{e}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (B) \(\mathrm{RT} \log _{10}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (C) RT \(\log _{10}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\) (D) RT \(\log _{\mathrm{e}}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\)

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