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Chapter 9: Problem 1282

The molecules of a given mass of a gas have a rms velocity of $200 \mathrm{~m} / \mathrm{s}\( at \)27^{\circ} \mathrm{C}\( and \)1.0 \times 10^{5} \mathrm{Nm}^{-2}\( pressure when the temperature is \)127^{\circ} \mathrm{C}$ and pressure is \(0.5 \times 10^{5} \mathrm{Nm}^{-2}\), the rms velocity in \(\mathrm{m} / \mathrm{s}\) will be (A) \(100 \sqrt{2}\) (B) \([(100 \sqrt{2}) / 3]\) (C) \([(400) / \sqrt{3}]\) (D) None of these

Short Answer

Expert verified
The new rms velocity when the temperature is 127°C and the pressure is 0.5 x 10^5 Nm^-2 will be approximately \(100 \sqrt{2} m/s\). Therefore, the correct answer is (A) \(100 \sqrt{2}\).

Step by step solution

01

Write down the given values

We are given: - Initial rms velocity: \(v_1 = 200 m/s\) - Initial temperature: \(T_1 = 27°C = 273 + 27 = 300 K\) - Initial pressure: \(P_1 = 1.0 \times 10^5 Nm^{-2}\) - Final temperature: \(T_2 = 127°C = 273 + 127 = 400 K\) - Final pressure: \(P_2 = 0.5 \times 10^5 Nm^{-2}\)
02

Write down the ideal gas law equation and rms velocity equation

The ideal gas law equation is: \( PV = nRT \) The rms velocity equation is: \( v_{rms} = \sqrt{\frac{3RT}{M}} \) where - P is the pressure. - V is the volume. - n is the number of gas molecules. - R is the universal gas constant. - T is the temperature. - M is the molar mass of the gas.
03

Calculate the final rms velocity

We can combine the ideal gas law equation and the rms velocity equation as follows: \( \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1} \cdot \frac{P_1}{P_2}} \) Substitute the given values into the equation: \( \frac{v_2}{200} = \sqrt{\frac{400}{300} \cdot \frac{1.0 \times 10^5}{0.5 \times 10^5}} \) Now, solve for \(v_2\): \( v_2 = 200 \cdot \sqrt{\frac{4}{3} \cdot 2} \) \( v_2 = 200 \cdot \sqrt{\frac{8}{3}} \) \( v_2 \approx 100 \sqrt{2} (m/s) \) The new rms velocity when the temperature is 127°C and the pressure is 0.5 x 10^5 Nm^-2 will be approximately \(100 \sqrt{2} m/s\). Therefore, the correct answer is (A) \(100 \sqrt{2}\).

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Most popular questions from this chapter

The average kinetic energy of a gas molecule at \(27^{\circ} \mathrm{C}\) is \(6.21 \times 10^{-21} \mathrm{~J}\). Its average kinetic energy at $227^{\circ} \mathrm{C}$ will be (A) \(5.22 \times 10^{-21} \mathrm{~J}\) (B) \(11.35 \times 10^{-21} \mathrm{~J}\) (C) \(52.2 \times 10^{-21} \mathrm{~J}\) (D) \(12.42 \times 10^{-21} \mathrm{~J}\)

Suppose ideal gas equation follows \(V P^{3}=\) constant, Initial temperature and volume of the gas are \(\mathrm{T}\) and \(\mathrm{V}\) respectively. If gas expand to \(27 \mathrm{~V}\), then temperature will become (A) \(9 \mathrm{~T}\) (B) \(27 \mathrm{~T}\) (C) (T/9) (D) \(\mathrm{T}\)

Hydrogen gas is filled in a balloon at \(20^{\circ} \mathrm{C}\). If temperature is made \(40^{\circ} \mathrm{C}\), pressure remaining the same what fraction of hydrogen will come out (A) \(0.75\) (B) \(0.07\) (C) \(0.25\) D) \(0.5\)

At \(100 \mathrm{~K}\) and \(0.1\) atmospheric pressure, the volume helium gas is 10 liters. If volume and pressure are doubled, its temperature will change to (A) \(127 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) (C) \(25 \mathrm{~K}\) (D) \(200 \mathrm{~K}\)

The pressure is exerted by the gas on the walls of the container because (A) It sticks with the walls (B) It is accelerated towards the walls (C) It loses kinetic energy (D) On collision with the walls there is a change in momentum
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