Question

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth is closest to:
[Boltzmans Constant ᵏB = 1.38 10⁻²³ J/K Avogadro number Nₐ = 6.02 × 10²⁶ /kg
Radius of Earth: 6.4 × 10⁶ m
Gravitation acceleration on Earth = 10 ms⁻²
] (A) 800k (B) 10⁴ K
(C) 3 ×10⁵ (D) 650 K

Answer 1

B) 10⁴ K

The rms velocity of Hydrogen will be equal to their escape velocity from Earth at a temperature of 10⁴ K

• Given :

Boltzmann Constant K$_B$ =  1.38 × 10⁻²³J/K

Avogadro number Nₐ = 6.02 × 10²⁶ /kg

Radius of Earth: R$_e$ = 6.4 × 10⁶ m

Gravitation acceleration on Earth = g = 10 ms⁻²

• Gas constant (R) is given by product of Boltzmann constant and the Avogadro number.

$R= K_{B} \times N_{A} = 1.38\times 10^{-23} \times 6.02 \times 10^{26} \\\\R = 8.3076 \times 10^{3}\ JK^{-1}kg^{-1}$........................(1)

• Mass of hydrogen molecule in kg can be given as 2 kg. (Since Avogadro number is given in units of /kg)

M = 2 kg ........................(2)

• The root mean square velocity of hydrogen molecule is given by

$Vrms = \sqrt{\frac{3RT}{M}}$  .......................(3)

• The escape velocity of Earth is given by

$V_{esc} = \sqrt{2gR_e}$  ..............................(4)

• Given condition states that root mean square velocity of hydrogen is equal to their escape velocity from Earth. Equating (3) and (4), we get

$V_{rms} = V_{escape}\\\\\sqrt{\frac{3RT}{M}} = \sqrt{2gR_e}\\\\\frac{3RT}{M}} = 2gR_e\\\\T = \frac{2gR_e M}{3R}$ ..............................................(5)

Substituting the values in (5), we get

$T = \frac{2\times 10\times 6.4\times 10^6\times 2}{3\times 8.3076\times 10^3} = \frac{ 256\times 10^3}{24.9228}\\\\T = 10.2 \times 10^3$

T ≈ 10 × 10³ = 10⁴ Kelvin

Answer 2

Thus the value of temperature is T = 10^4 K.  

Option (B) is correct.

Explanation:

Given data:

• Boltzmann Constant K_B =  1.38 × 10⁻²³J/K
• Avogadro number Nₐ = 6.02 × 10²⁶ /kg
• Radius of Earth: R_e = 6.4 × 10⁶ m
• Gravitation acceleration on Earth = g = 10 ms⁻²

vrms =  √2RT / m , Vescape  =  √2gRe

Vrms = Vescape

3 RT / m = 2gRe

3 x 1.38 x 10^-23 x 6.02 x 10^26/ 2 x T

T = 2 x 10 x 6.4 x 10^6

T = 4 x 10 x 6.4 x 10^6 / 3 x 1.38 x 6.02 x 10^3  = 10 x 10^3

T = 10^4 K

Thus the value of temperature is T = 10^4 K.