The temperature, at which the root mean square velocity of hydrogen mo

1.	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 /kgRadius of Earth: 6.4 × 10⁶ mGravitation acceleration on Earth = 10 ms⁻²] (A) 800k (B) 10⁴ K(C) 3 ×10⁵ (D) 650 K
Answer» 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

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 /kgRadius of Earth: 6.4 × 10⁶ mGravitation acceleration on Earth = 10 ms⁻²] (A) 800k (B) 10⁴ K(C) 3 ×10⁵ (D) 650 K

Answer»

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

B) 10⁴ K

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