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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2021
  • Hard
  • MCQ

For an ideal gas the instantaneous change in pressure \('{p} '\) with volume \('v'\) is given by the equation \(\frac{{dp}}{{dv}}=-{ap} .\) If \({p}={p}_{0}\) at \({v}=0\) is the given boundary condition, then the maximum temperature one mole of gas can attain is : (Here \({R}\) is the gas constant)

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Answer & Solution

Correct Answer

(A) \(\frac{{p}_{0}}{{aeR}}\)

Step-by-step Solution

Detailed explanation

\(\int_{p_{0}}^{p} \frac{d p}{P}=-a \int_{0}^{v} d v\)

\(\ell n\left(\frac{p}{p_{0}}\right)=-a v\)

\(p=p_{0} e^{-a v}\)

For temperature maximum p-v product should be maximum

\({T}=\frac{{pv}}{{n} {R}}=\frac{{p}_{0} {ve}^{-{av}}}{{R}}\)

\(\frac{{dT}}{{d} {v}}=0 \Rightarrow \frac{{p}_{0}}{{R}}\left\{{e}^{-{av}}+{ve}^{-{dv}}(-{a})\right\}\)

\(\frac{{p}_{0} {e}^{-{dv}}}{{R}}\{1-{av}\}=0\)

\({v}=\frac{1}{{a}}, \infty\)

\({T}=\frac{{p}_{0} 1}{{Rae}}=\frac{{p}_{0}}{{Rae}}\)

\(\text { at } {v}=\infty\)

\({T}=0\)

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