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Chapter 8: Problem 1115

An insulated contains containing monoatomic gas of moles mass \(\mathrm{M}_{0}\) is moving with a velocity, \(\mathrm{V}\). If the container is suddenly stopped, find the change in temperature. (A) \(\left\\{\left(M_{0} V^{2}\right) /(5 R)\right\\}\) (B) $\left\\{\left(\mathrm{M}_{0} \mathrm{~V}^{2}\right) /(4 \mathrm{R})\right\\}$ (C) $\left\\{\left(\mathrm{M}_{0} \mathrm{~V}^{2}\right) /(3 \mathrm{R})\right\\}$ (D) $\left\\{\left(\mathrm{M}_{0} \mathrm{~V}^{2}\right) /(2 \mathrm{R})\right\\}$

Short Answer

Expert verified
The change in temperature is \(\Delta T = \frac{M_0 \times V^2}{3R}\). The correct option is (C).

Step by step solution

01

Write down the conservation of energy formula

We need to apply conservation of energy, so we will equate the initial kinetic energy of the container to the final change in internal energy of the gas: \( E_{kinetic} = \Delta E_{internal} \)
02

Write down initial kinetic energy

The initial kinetic energy is given by: \( E_{kinetic} = \frac{1}{2} M_0 V^2 \)
03

Write down internal energy change

For a monoatomic gas, the change in internal energy ΔE is given by: \( \Delta E_{internal} = \frac{3}{2} n R \Delta T \) where n is the number of moles and R is the gas constant.
04

Calculate the number of moles

We need to find the number of moles (n) from the given molar mass M₀: \( n = \frac{M_0}{Molar\:mass} \)
05

Apply conservation of energy

Now we will apply conservation of energy using initial kinetic energy and change in internal energy. Replace ΔE with the appropriate terms: \( \frac{1}{2} M_0 V^2 = \frac{3}{2} n R \Delta T \) Replace n: \( \frac{1}{2} M_0 V^2 = \frac{3}{2} \frac{M_0}{Molar\:mass} R \Delta T \) Now we need to solve for \(\Delta T\):
06

Solve for the change in temperature

To solve for \(\Delta T\), follow these steps: Divide both sides by M₀, and then multiply both sides by 2: \( V^2 = 3 \frac{R}{Molar\:mass} \Delta T \) Now, multiply both sides by Molar mass and divide by 3R: \( \Delta T = \frac{Molar\:mass \times V^2}{3R} \) Since we are given initial molar mass M₀, and we know that for an ideal gas, Molar mass = M₀: \( \Delta T = \frac{M_0 \times V^2}{3R} \) So, the change in temperature is given by the formula: \( \Delta T = \left\\{ \left( M_0 V^2 \right) / \left( 3R \right) \right\\} \) Thus, the answer is (C).

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Most popular questions from this chapter

If du represents the increase in internal energy of a thermodynamic system and dw the work done by the system, which of the following statement is true? (A) \(\mathrm{du}=\mathrm{dw}\) in isothermal process (C) \(\mathrm{du}=-\mathrm{dw}\) in an aidabatic process (B) \(\mathrm{du}=\mathrm{dw}\) in aidabatic process (D) \(\mathrm{du}=-\mathrm{dw}\) in an isothermal process

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 \(\log _{\mathrm{e}}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\)

70 calorie of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from \(30^{\circ} \mathrm{C}\) to $35^{\circ} \mathrm{C}$ The amount of heat required to raise the temperature of the same gas through the same range at constant volume is $\ldots \ldots \ldots \ldots \ldots .$ calorie. (A) 50 (B) 30 (C) 70 (D) 90

One mole of a monoatomic gas \([\gamma=(5 / 3)]\) is mixed with one mole of A diatomic gas \([\gamma=(7 / 5)]\) what will be value of \(\gamma\) for mixture ? (A) \(1.454\) (B) \(1.4\) (C) \(1.54\) (D) \(1.5\)

An ideal refrigerator has a freezer at a temperature of \(-13\) C, The coefficient of performance of the engine is 5 . The temperature of the air to which heat is rejected will be. (A) \(325^{\circ} \mathrm{C}\) (B) \(39^{\circ} \mathrm{C}\) (C) \(325 \mathrm{~K}\) (D) \(320^{\circ} \mathrm{C}\)
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