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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)
Pick an option to check your answer.
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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