English
Karnataka Board PUCPUC Science Class 11

Two molecules of a gas have speeds of 9 × 10 6 ms−1 and 1 × 106 ms−1, respectively. What is the root mean square speed of these molecules?

Advertisements
Advertisements

Question

Two molecules of a gas have speeds of 9 × 10 6 ms−1 and 1 × 106 ms−1, respectively. What is the root mean square speed of these molecules?

Short/Brief Note
Advertisements

Solution

For n-molecules, we know that

`v_(rms) = sqrt((v_1^2 + v_2^2 + v_3^2 + ...... + v_n^2)/n`  .....`[(v_(rms) = "root mean"),("square velocity")]`

Where v1, v2, v3 ....... vn are individual velocities of n-molecules of the gas.

For two molecules,

`v_(rms) = sqrt((v_1^2 + v_2^2)/2`  ......[v1, v2, v3 ....... vn are individual velocity]

Given, `v_1 = 9 xx 10^6` m/s

And `v_2 = 1 xx 10^6` m/s

∴ `v_(rms) = sqrt(((9 xx 10^6)^2 + (1 xx 10^6)^2)/2`

= `sqrt((81 xx 10^12 + 1 xx 10^12)/2`

= `sqrt(((81 + 1) xx 10^12)/2`

= `sqrt((82 xx 10^12)/2`

= `sqrt(41) xx 10^6` m/s

shaalaa.com
  Is there an error in this question or solution?
Chapter 13: Kinetic Theory - Exercises [Page 94]

APPEARS IN

NCERT Exemplar Physics Exemplar [English] Class 11
Chapter 13 Kinetic Theory
Exercises | Q 13.17 | Page 94

RELATED QUESTIONS

Comment on the following statement: the temperature of all the molecules in a sample of a gas is the same.


It is said that the assumptions of kinetic theory are good for gases having low densities. Suppose a container is so evacuated that only one molecule is left in it. Which of the assumptions of kinetic theory will not be valid for such a situation? Can we assign a temperature to this gas?


If the molecules were not allowed to collide among themselves, would you expect more evaporation or less evaporation?


The mean speed of the molecules of a hydrogen sample equals the mean speed of the molecules of a helium sample. Calculate the ratio of the temperature of the hydrogen sample to the temperature of the helium sample.

Use R = 8.314 JK-1 mol-1


Figure shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are pA, TA, V in the vessel A and pB, TB, V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy `Ρ/T = 1/2 ({P_A}/{T_A}+{P_B}/{T_B))` when equilibrium is achieved.


Figure shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction coefficient between the cork and the tube is 0.20. The tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 K. The tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. Let dN denote the magnitude of the normal contact force exerted by a small length dlof the cork along the periphery (see the figure). Assuming that the temperature of the gas is uniform at any instant, calculate `(dN)/(dt)`.


The condition of air in a closed room is described as follows. Temperature = 25°C, relative humidity = 60%, pressure = 104 kPa. If all the water vapour is removed from the room without changing the temperature, what will be the new pressure? The saturation vapour pressure at 25°C − 3.2 kPa.


Figure shows two rigid vessels A and B, each of volume 200 cm3, containing an ideal gas (Cv = 12.5 J K−1 mol−1). The vessels are connected to a manometer tube containing mercury. The pressure in both the vessels is 75 cm of mercury and the temperature is 300 K. (a) Find the number of moles of the gas in each vessel. (b) 5.0 J of heat is supplied to the gas in vessel A and 10 J to the gas in vessel B. Assuming there's no appreciable transfer of heat from A to B, calculate the difference in the heights of mercury in the two sides of the manometer. Gas constant, R = 8.3 J K−1 mol−1.


Using figure, find the boiling point of methyl alcohol at 1 atm (760 mm of mercury) and at 0.5 atm.


Calculate the average molecular kinetic energy 

  1. per kmol 
  2. per kg 
  3. per molecule 

of oxygen at 127°C, given that the molecular weight of oxygen is 32, R is 8.31 J mol−1K1 and Avogadro’s number NA is 6.02 × 1023 molecules mol1.


Find the temperature of a blackbody if its spectrum has a peak at (a) λmax = 700 nm (visible), (b) λmax = 3 cm (microwave region) (c) λmax = 3 m (short radio waves). (Take Wien’s constant b = 2.897 × 10-3 m.K).


Calculate the value of λmax for radiation from a body having a surface temperature of 3000 K. (b = 2.897 x 10-3 m K) 


Calculate the energy radiated in one minute by a blackbody of surface area 200 cm2 at 127 °C (σ = 5.7 x 10-8 J m-2 s-1 K-4)  


Why the temperature of all bodies remains constant at room temperature?


The graph of kinetic energy against the frequency v of incident light is as shown in the figure. The slope of the graph and intercept on X-axis respectively are ______.


The average translational kinetic energy of a molecule in a gas becomes equal to 0.49 eV at a temperature about (Boltzmann constant = 1.38 x 10-23 JK-1) ____________.


An inflated rubber balloon contains one mole of an ideal gas, has a pressure p, volume V and temperature T. If the temperature rises to 1.1 T, and the volume is increased to 1.05 V, the final pressure will be ______.


For a particle moving in vertical circle, the total energy at different positions along the path ______.


When a particle oscillates simple harmonically, its kinetic energy varies periodically. If frequency of the particle is n, then the frequency of the kinetic energy is ______.


According to the kinetic theory of gases, at a given temperature, molecules of all gases have the same ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×