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Chapter 9: Problem 1291

At what temperature the molecules of nitrogen will have the same rms. velocity as the molecules of Oxygen at \(127^{\circ} \mathrm{C}\). (A) \(273^{\circ} \mathrm{C}\) (B) \(350^{\circ} \mathrm{C}\) (C) \(77^{\circ} \mathrm{C}\) (D) \(457^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperature at which the molecules of Nitrogen will have the same rms velocity as the molecules of Oxygen at 127°C is \(T_{N_2} = 77^{\circ} \mathrm{C}\) (Option C).

Step by step solution

01

Understanding rms velocity formula

We need to first recall the formula for rms velocity, which is: \(v_{rms} = \sqrt{\frac{3RT}{M}} \) where: - \(v_{rms}\) is the root-mean-square velocity, - R is the universal gas constant (8.31 J/(mol K)), - T is the temperature in Kelvin, and - M is the molar mass of the given gas in kg/mol.
02

Converting given temperature to Kelvin

We are given that the temperature of the Oxygen gas is 127°C. To use the rms velocity formula, we need to convert this temperature to Kelvin: (127 + 273) K = 400 K
03

Calculate rms velocity of Oxygen

Using the rms velocity formula, we calculate the rms velocity of Oxygen at 127°C (400 K). The molar mass of Oxygen (O₂) is 32 g/mol or 0.032 kg/mol. \(v_{O_2} = \sqrt{\frac{3 × 8.31 × 400}{0.032}}\)
04

Setting up an equation for Nitrogen's temperature

To find the temperature at which Nitrogen has the same rms velocity as Oxygen, we set up an equation: \(v_{O_2} = v_{N_2}\) We know that the molar mass of Nitrogen (N₂) is 28 g/mol or 0.028 kg/mol. Using the rms velocity formula for both gases and setting their velocities equal, we get: \(\sqrt{\frac{3 × 8.31 × 400}{0.032}} = \sqrt{\frac{3 × 8.31 × T_{N_2}}{0.028}}\)
05

Solve the equation for Nitrogen's temperature

Now we solve the equation for \(T_{N_2}\): \(\frac{3 × 8.31 × 400}{0.032} = \frac{3 × 8.31 × T_{N_2}}{0.028}\) \(T_{N_2} = \frac{400 × 0.028}{0.032}\) \(T_{N_2} = 350\, \text{K}\)
06

Convert Nitrogen's temperature to Celsius

Now we convert the temperature 350 K back to Celsius: (350 - 273)°C = 77°C Thus, we find that the temperature at which Nitrogen has the same rms velocity as Oxygen at 127°C is: \(T_{N_2} = 77^{\circ} \mathrm{C}\) The correct option is (C) 77°C.

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Most popular questions from this chapter

At what temperature pressure remaining constant will the \(\mathrm{rms}\) speed of a gas molecules increase by \(10 \%\) of the \(\mathrm{rms}\) speed at NTP? (A) \(57.3 \mathrm{~K}\) (B) \(57.3^{\circ} \mathrm{C}\) (C) \(557.3 \mathrm{~K}\) (D) \(-57.3^{\circ} \mathrm{C}\)

1 mole of gas occupies a volume of \(100 \mathrm{~m} 1\) at \(50 \mathrm{~mm}\) pressure. What is the volume occupied by two moles of gas at $100 \mathrm{~mm}$ pressure and at same temperature (A) \(50 \mathrm{ml}\) (B) \(200 \mathrm{ml}\) (C) \(100 \mathrm{ml}\) (D) \(500 \mathrm{~m} 1\)

A gas at \(27^{\circ} \mathrm{C}\) has a volume \(\mathrm{V}\) and pressure \(\mathrm{P}\). On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be (A) \(1527^{\circ} \mathrm{C}\) (B) \(600^{\circ} \mathrm{C}\) (C) \(162^{\circ} \mathrm{C}\) (D) \(1800^{\circ} \mathrm{C}\)

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

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) \(0.5\)
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