Advertisements
Advertisements
प्रश्न
The molecules of a given mass of a gas have root mean square speeds of 100 ms−1 at 27°C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?
उत्तर
We know that for a given mass of a gas
`V_(rms) = sqrt((3RT)/M)`
Where R is gas constant
T is the temperature in Kelvin
M is the molar mass of the gas
Clearly, `V_(rms) ∝ sqrt(T)`
As R and M are constants,
`(V_(rms))_1/(V_(rms))_2 = sqrt(T_1/T_2)`
Given, `(V_(rms))_1` = 100 m/s
T1 = 27°C = 27 + 273 = 300 K
T2 = 127°C = 127 + 273 = 400 K
∴ From equation (i)
`100/((V_(rms))_2) = sqrt(300/400) = sqrt(3)/2`
⇒ `(V_(rms))_2 = (2 xx 100)/sqrt(3) = 200/sqrt(3)` m/s
APPEARS IN
संबंधित प्रश्न
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
Answer in brief:
What will happen to the mean square speed of the molecules of a gas if the temperature of the gas increases?
Compare the rates of emission of heat by a blackbody maintained at 727°C and at 227°C, if the black bodies are surrounded by an enclosure (black) at 27°C. What would be the ratio of their rates of loss of heat?
The power radiated by a perfect blackbody depends only on its ______
Under which condition laws of Boyle, Charles, and Gay-Lussac are valid?
Why the temperature of all bodies remains constant at room temperature?
Above what temperature, all bodies radiate electromagnetic radiation?
A gas mixture consists of molecules of types A, B and C with masses mA > mB > mC. Rank the three types of molecules in decreasing order of rms speeds.
Assuming the expression for the pressure P exerted by an ideal gas, prove that the kinetic energy per unit volume of the gas is `3/2` P.
According to the kinetic theory of gases, at a given temperature, molecules of all gases have the same ______.
