JEE MAIN - Physics (2021 - 31st August Morning Shift - No. 13)
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)
$${{{p_0}} \over {aeR}}$$
$${{a{p_0}} \over {eR}}$$
infinity
0$$^\circ$$C
Explanation
$$\int\limits_{{p_0}}^p {{{dp} \over P} = - a\int\limits_0^v {dv} } $$
$$\ln \left( {{p \over {{p_0}}}} \right) = - av$$
$$p = {p_0}{e^{ - av}}$$
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)
$$\ln \left( {{p \over {{p_0}}}} \right) = - av$$
$$p = {p_0}{e^{ - av}}$$
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)
Comments (0)
