Question 561: Thermodynamics & KTG

Question

The root mean square speed of molecules of a given mass of a gas at 27$$^\circ$$C and 1 atmosphere pressure is 200 ms
$$-$$1
. The root mean square speed of molecules of the gas at 127$$^\circ$$C and 2 atmosphere pressure is $${{x \over {\sqrt 3 }}}$$ ms
$$-$$1
. The value of x will be _________.

Accepted Answer

Given, T
1
= 27$$^\circ$$C = 27 + 273 = 300K, p
1
= 1 atm, v
1
= 200 ms
$$-$$1
, T
2
= 127$$^\circ$$C = 400 K, p
2
= 12 atm, v
2
= ?
As we know that,
Root mean square speed, $${v_{rms}} = \sqrt {{{3RT} \over m}} $$
$$	herefore$$ $${{{v_1}} \over {{v_2}}} = \sqrt {{{{T_1}} \over {{T_2}}}}  = \sqrt {{{300} \over {400}}}  = \sqrt {{3 \over 4}} $$
$$ \Rightarrow {v_2} = \sqrt {{4 \over 3}} {v_1} = {2 \over {\sqrt 3 }} 	imes 200 = {{400} \over {\sqrt 3 }}$$ ms
$$-$$1
$$ \Rightarrow {x \over {\sqrt 3 }} = {{400} \over {\sqrt 3 }} \Rightarrow x = 400$$

Thermodynamics & KTG

The root mean square speed of molecules of a given mass of a gas at 27$^\circC and 1 atmosphere pressure is 200 ms -1 . The root mean square speed of molecules of the gas at 127^\circC and 2 atmosphere pressure is {{x \over {\sqrt 3 }}} ms -$1 . The value of x will be _________.

Integer / Numeric Answer Type