The temperature of an ideal gas is increased from 100 K to 400 K. If the rms speed of the gas molecule is v at 100 K, then at 400 K it becomes

Question

The temperature of an ideal gas is increased from 100 K to 400 K. If the rms speed of the gas molecule is v at 100 K, then at 400 K it becomes (A) 4v (B) 2v (C) 0.5v (D) 0.25v

Tardigrade Admin · Accepted Answer

∵ rms speed of molecule V text rms propto √T for T=100 K (V rms )1 propto √100 for T=400 K (V rms )2=√400 therefore, ((V rms )1/(V rms )2)=(√100/√400) or (v/(V rms )2)=(1/2) or (V rms )2=2 v

Q. The temperature of an ideal gas is increased from $100 K$ to $400 K$ . If the rms speed of the gas molecule is $v$ at $100 K$ , then at $400 K$ it becomes

4v

2v

0.5v

0.25v

v

$∵$ rms speed of molecule $V_{rms} \propto T$
for $T = 100 K$
$(V_{r m s})_{1} \propto 100$
for $T = 400 K$
$(V_{r m s})_{2} = 400$
therefore, $\frac{( V _{r m s} ) _{1}}{( V _{r m s} ) _{2}} = \frac{100}{400}$
or $\frac{v}{( V _{r m s} ) _{2}} = \frac{1}{2}$
or $(V_{r m s})_{2} = 2 v$

Q. The temperature of an ideal gas is increased from 100K to 400K. If the rms speed of the gas molecule is v at 100K, then at 400K it becomes

4v

2v

0.5v

0.25v

v

∵ rms speed of molecule Vrms ​∝T​ for T=100K (Vrms​)1​∝100​ for T=400K (Vrms​)2​=400​ therefore, (Vrms​)2​(Vrms​)1​​=400​100​​ or (Vrms​)2​v​=21​ or (Vrms​)2​=2v

Q. The temperature of an ideal gas is increased from $100 K$ to $400 K$ . If the rms speed of the gas molecule is $v$ at $100 K$ , then at $400 K$ it becomes

$∵$ rms speed of molecule $V_{rms} \propto T$
for $T = 100 K$
$(V_{r m s})_{1} \propto 100$
for $T = 400 K$
$(V_{r m s})_{2} = 400$
therefore, $\frac{( V _{r m s} ) _{1}}{( V _{r m s} ) _{2}} = \frac{100}{400}$
or $\frac{v}{( V _{r m s} ) _{2}} = \frac{1}{2}$
or $(V_{r m s})_{2} = 2 v$