JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 15)
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
$${3 \over 4}{{{P_0}{V_0}} \over R}$$
$${1 \over 2}{{{P_0}{V_0}} \over R}$$
$${5 \over 4}{{{P_0}{V_0}} \over R}$$
$${1 \over 4}{{{P_0}{V_0}} \over R}$$
Explanation
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 Ti = $${{{V_0}} \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over {{V_0}}}} \right)}^2}} \right]$$
= $${{{V_0}{P_0}} \over {2R}}$$
Case 2 : when V = 2V0
then Tf = $${{2{V_0}} \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over {2{V_0}}}} \right)}^2}} \right]$$
= $${7 \over 4}{{{P_0}{V_0}} \over R}$$
Then $$\Delta $$T = Tf - Ti
= $${7 \over 4}{{{P_0}{V_0}} \over R} - {{{P_0}{V_0}} \over {2R}}$$
= $${{{P_0}{V_0}} \over R}\left( {{7 \over 4} - {1 \over 2}} \right)$$
= $${5 \over 4}{{{P_0}{V_0}} \over R}$$
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 Ti = $${{{V_0}} \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over {{V_0}}}} \right)}^2}} \right]$$
= $${{{V_0}{P_0}} \over {2R}}$$
Case 2 : when V = 2V0
then Tf = $${{2{V_0}} \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over {2{V_0}}}} \right)}^2}} \right]$$
= $${7 \over 4}{{{P_0}{V_0}} \over R}$$
Then $$\Delta $$T = Tf - Ti
= $${7 \over 4}{{{P_0}{V_0}} \over R} - {{{P_0}{V_0}} \over {2R}}$$
= $${{{P_0}{V_0}} \over R}\left( {{7 \over 4} - {1 \over 2}} \right)$$
= $${5 \over 4}{{{P_0}{V_0}} \over R}$$
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