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Heat and Thermodynamics question

2024 · 9 Apr · Shift 2 · Q74
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Heat and Thermodynamics question

2024 · 9 Apr · Shift 2 · Q74

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The temperature of a gas is −78∘C-78^{\circ} \mathrm{C}−78∘C and the average translational kinetic energy of its molecules is K\mathrm{K}K. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes 2 K2 \mathrm{~K}2 K is :
  1. A
    −78∘C-78^{\circ} \mathrm{C}−78∘C
  2. B
    127∘C127^{\circ} \mathrm{C}127∘C
  3. C
    −39∘C-39^{\circ} \mathrm{C}−39∘C
  4. D
    117∘C117^{\circ} \mathrm{C}117∘C
View written solutionFree

Correct answer: D

  1. Use the relation between average translational kinetic energy and absolute temperature

For an ideal gas, the average translational kinetic energy per molecule is

Eˉ=32kT\bar{E} = \frac{3}{2}kTEˉ=23​kT

So, average kinetic energy is directly proportional to the absolute temperature TTT (in kelvin).

  1. Convert the given temperature to kelvin

Given temperature:

−78∘C-78^\circ \text{C}−78∘C

In kelvin,

T1=273+(−78)=195 KT_1 = 273 + (-78) = 195\,\text{K}T1​=273+(−78)=195K

At this temperature, the average kinetic energy is KKK.

  1. For kinetic energy to become 2K2K2K

Since

Eˉ∝T\bar{E} \propto TEˉ∝T

if the kinetic energy doubles, the absolute temperature must also double:

T2=2T1=2×195=390 KT_2 = 2T_1 = 2 \times 195 = 390\,\text{K}T2​=2T1​=2×195=390K
  1. Convert back to degree Celsius
T2=390−273=117∘CT_2 = 390 - 273 = 117^\circ \text{C}T2​=390−273=117∘C
  1. Match with the options
117∘C117^\circ \text{C}117∘C

So the correct option is D.

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