QJEEM19Ph6a54: JEE Main 2019 Physics - Class 11 JEEM19Ph6a54 Solution: One mole of ideal gas passes through a process where pressure…

Question

Verified Solution • 2025–26 Session moderate

One mole of ideal gas passes through a process where pressure and volume obey the relation

P = {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over V}} \right)}^2}} \right]

. Here P₀ and V₀ are constants. Calculate the change in the temperature of the gas if its volume changes form V₀ to 2V₀

Accepted Answer

Given

P = {P_o}\left\{ {1 - {1 \over 2}{{\left( {{{{V_o}} \over V}} \right)}^2}} \right\};

...(i)

As n = 1 mole

\therefore

PV = nRT = RT

\Rightarrow

P =

{{RT} \over V}

....(ii)

From (i) and (ii), we get

{{RT} \over V} = {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over V}} \right)}^2}} \right]

\Rightarrow

T =

{V \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over V}} \right)}^2}} \right]

Case 1 : when V = V₀

then T_i =

{{{V_0}} \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over {{V_0}}}} \right)}^2}} \right]

=

{{{V_0}{P_0}} \over {2R}}

Case 2 : when V = 2V₀

then T_f =

{{2{V_0}} \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over {2{V_0}}}} \right)}^2}} \right]

=

{7 \over 4}{{{P_0}{V_0}} \over R}

Then

\Delta

T = T_f - T_i

=

{7 \over 4}{{{P_0}{V_0}} \over R} - {{{P_0}{V_0}} \over {2R}}

=

{{{P_0}{V_0}} \over R}\left( {{7 \over 4} - {1 \over 2}} \right)

=

{5 \over 4}{{{P_0}{V_0}} \over R}

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