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JEE Main 20231 FebruaryEvening Shift
Exam
JEE Main
Year
2023
Date
1 February
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2023 · 1 February · Evening Shift · Physics · Thermodynamics

For three low density gases A, B, C pressure versus temperature...

For three low density gases A,B,CA, B, C pressure versus temperature graphs are plotted while keeping them at constant volume, as shown in the figure

For three low density gases A, B, C pressure versus temperature graphs are plotted while keeping them at constant

The temperature corresponding to the point ′K′' K ' is:

Options

  1. A−273°C- 273 ° C
  2. B−100°C- 100 ° C
  3. C–373°C– 373 ° C
  4. D−40°C- 40 ° C

Correct answer

A. −273°C- 273 ° C

Step-by-step solution

In case of constant volume process, i.e., isochoric process,

From ideal gas equation,

PT=nRV=\dfrac{P}{T} = n\dfrac{R}{V} =constant

Or we have, P=nRV(t+273)P = \dfrac{nR}{V}\left(t + 273\right)

If P=0⇒t=−273°CP = 0 ⇒ t = - 273 ° C

Thus, when temperature goes to −273°C- 273 ° C, all constituent particles get freezed and pressure goes to zero irrespective of the density of gas.

Hence, the temperature corresponding to the point K=−273°CK = - 273 ° C.

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