"For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation ${{dp} \over {dv}} = - ap$. If p = p0 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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Accepted Answer

"$\int\limits_{{p_0}}^p {{{dp} \over P} = - a\int\limits_0^v {dv} }$

$\ln \left( {{p \over {{p_0}}}} \right) = - av$

$p = {p_0}{e^{ - av}}$

For temperature maximum p-v product should be maximum

$T = {{pv} \over {nR}} = {{{p_0}v{e^{ - av}}} \over R}$

$${{dT} \over {dv}} = 0 \Rightarrow {{{p_0}} \over R}\{ {e^{ - av}} + v{e^{ - av}}( - a)\} $$ = 0

${{{p_0}{e^{ - av}}} \over R}\{ 1 - av\} = 0$

$v = {1 \over a},\infty$

$T = {{{p_0}1} \over {Rae}} = {{{p_0}} \over {Rae}}$

at v = $\infty$

T = 0

Option (a)"

For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation ${{dp} \over {dv}} = - ap$. If p = p₀ 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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For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation ${{dp} \over {dv}} = - ap$. If p = p0 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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For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation ${{dp} \over {dv}} = - ap$. If p = p₀ 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)