JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 16)
The temperature, at which the root mean square
velocity of hydrogen molecules equals their
escape velocity from the earth, is closest to :
[Boltzmann Constant kB = 1.38 × 10–23 J/K Avogadro Number NA = 6.02 × 1026 /kg Radius of Earth : 6.4 × 106 m Gravitational acceleration on Earth = 10ms–2]
[Boltzmann Constant kB = 1.38 × 10–23 J/K Avogadro Number NA = 6.02 × 1026 /kg Radius of Earth : 6.4 × 106 m Gravitational acceleration on Earth = 10ms–2]
3 × 105 K
104 K
650 K
800 K
Explanation
$${V_{rms}} = \sqrt {{{3RT} \over M}} = 11.2 \times {10^3}m/s$$
$$ \Rightarrow $$ $$T = {M \over {3R}} \times {\left( {11.2 \times {{10}^3}} \right)^2}$$
= $${{2 \times {{10}^{ - 3}}} \over {3 \times 8.3}} \times 125.44 \times {10^6} = {10^4}K$$
$$ \Rightarrow $$ $$T = {M \over {3R}} \times {\left( {11.2 \times {{10}^3}} \right)^2}$$
= $${{2 \times {{10}^{ - 3}}} \over {3 \times 8.3}} \times 125.44 \times {10^6} = {10^4}K$$
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