JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 16)
Number of molecules in a volume of 4 cm3 of
a perfect monoatomic gas at some temperature
T and at a pressure of 2 cm of mercury is close
to?
(Given, mean kinetic energy of a molecule
(at T) is 4 $$ \times $$ 10–14 erg, g = 980 cm/s2, density of
mercury = 13.6 g/cm3)
(Given, mean kinetic energy of a molecule
(at T) is 4 $$ \times $$ 10–14 erg, g = 980 cm/s2, density of
mercury = 13.6 g/cm3)
5.8 $$ \times $$ 1018
4.0 $$ \times $$ 1016
5.8 $$ \times $$ 1016
4.0 $$ \times $$ 1018
Explanation
$$E = {3 \over 2}kT \Rightarrow \left( {T = {{2E} \over {3k}}} \right),$$
Also $$\,PV = NkT$$
$$P = \rho gh,\,V = 4c{m^3}$$
$$ \therefore $$ ($$\rho gh$$)V = $$Nk \times {{2E} \over {3k}}$$
$$ \Rightarrow $$ $$13.6 \times {10^3} \times 9.8 \times 2 \times {10^{ - 2}} \times 4 \times {10^{ - 6}}$$
$$ = Nk \times {{2E} \over {3k}} = {{N \times 2} \over 3} \times 4 \times {10^{ - 14}} \times {10^{-7}}$$
$$ \Rightarrow $$ $$N = {{13.6 \times 19.6 \times 4 \times {{10}^{ - 5}} \times 3 \times 10} \over 8}$$
$$ \Rightarrow $$ $$N = 399.84 \times {10^{16}}$$
$$ = 3.99 \times {10^{18}}$$
$$ \Rightarrow $$ $$N = 4 \times {10^{18}}$$
Also $$\,PV = NkT$$
$$P = \rho gh,\,V = 4c{m^3}$$
$$ \therefore $$ ($$\rho gh$$)V = $$Nk \times {{2E} \over {3k}}$$
$$ \Rightarrow $$ $$13.6 \times {10^3} \times 9.8 \times 2 \times {10^{ - 2}} \times 4 \times {10^{ - 6}}$$
$$ = Nk \times {{2E} \over {3k}} = {{N \times 2} \over 3} \times 4 \times {10^{ - 14}} \times {10^{-7}}$$
$$ \Rightarrow $$ $$N = {{13.6 \times 19.6 \times 4 \times {{10}^{ - 5}} \times 3 \times 10} \over 8}$$
$$ \Rightarrow $$ $$N = 399.84 \times {10^{16}}$$
$$ = 3.99 \times {10^{18}}$$
$$ \Rightarrow $$ $$N = 4 \times {10^{18}}$$
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