Thermodynamics

JEE Main level, single correct, medium

100 of 190

Question 100 of 190

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

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

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Answer: (A) −273∘C-273^{\circ}C

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\Rightarrow t=-273^{\circ}C
Thus, when temperature goes to −273∘C-273^{\circ}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^{\circ}C.

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