At what temperature KE of a gas molecules is half of its value at 327- c -A- 13-5- CB- 150- CC- 27- CD- -123- C

The correct option is C 27^∘c KE=3/2RT KE1_1/KE_2=T_1/T_2 1/1/2=600/T_2⇒ T_2 =300 k or 27^∘c ...

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Standard XII

Physics

Specific Heat Capacity

At what tempe...

Question

At what temperature KE of a gas molecules is half of its value at $327^{\circ} c ?$

${13.5}^{\circ} c$

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$150^{\circ} c$

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$27^{\circ} c$

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$- 123^{\circ} c$

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Solution

The correct option is
C

$27^{\circ} c$
Kinetic energy of a gas molecule
E = 3/2kT
where k is Boltzmann's constant.
∴E∝T
$\frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{E}{\frac{E}{2}} = \frac{600}{T_{2}} \Rightarrow T_{2} = 300 k$ or $27^{\circ} c$

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At what temperature KE of a gas molecules is half of its value at 327∘c?

The correct option is C 27∘cKinetic energy of a gas moleculeE = 3/2kTwhere k is Boltzmann's constant.∴E∝TE1E2=T1T2 EE2=600T2⇒T2=300k or 27∘c

At what temperature KE of a gas molecules is half of its value at $327^{\circ} c ?$

The correct option is
C

$27^{\circ} c$
Kinetic energy of a gas molecule
E = 3/2kT
where k is Boltzmann's constant.
∴E∝T
$\frac{E_{1}}{E_{2}} = \frac{T_{1}}{T_{2}}$
$\frac{E}{\frac{E}{2}} = \frac{600}{T_{2}} \Rightarrow T_{2} = 300 k$ or $27^{\circ} c$