JEE Advance - Physics (2020 - Paper 2 Offline - No. 13)
A spherical bubble inside water has radius R. Take the pressure inside the bubble and the water pressure to be p0. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes (R $$-$$ a). For a << R the magnitude of the work done in the process in given by (4$$\pi$$p0Ra2)X, where X is a constant and $$\gamma$$ = Cp/Cv = 41/30. The value of X is ________.
คำตอบ
2.05
คำอธิบาย
As change in the kinetic energy of the surface of the bubble is zero.
$$ \therefore $$ Wexternal + Wwater + Wgas = 0 ....(i)
where Wexternal = work done by external agent
Wwater = work done by water
Wgas = work done by gas
For adiabatic process, $${P_1}V_1^\gamma = {P_2}V_2^\gamma $$
Here, $${P_0}{\left[ {{4 \over 3}\pi {R^3}} \right]^{{{41} \over {30}}}} = P{\left[ {{4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]^{{{41} \over {30}}}}$$
$$ \Rightarrow $$ P = $${P_0}{\left[ {{R \over {R - a}}} \right]^{{{41} \over {10}}}}$$
Now, Wgas = $${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$$
= $${{{P_0} \times {4 \over 3}\pi {R^3} - P \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$$
= $${{{P_0} \times {4 \over 3}\pi {R^3} - {P_0}{{\left[ {{R \over {R - a}}} \right]}^{{{41} \over {10}}}} \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$$
= $${{{P_0} \times {4 \over 3}\pi {R^3}\left[ {1 - {{\left( {{R \over {R - a}}} \right)}^{{{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]} \over {{{11} \over {30}}}}$$
= $${{{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {{{R - a} \over R}} \right)}^{ - {{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]}$$
= $${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {1 - {a \over R}} \right)}^{ - {{11} \over {10}}}}} \right]$$
= $${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - 1 - {{11a} \over {10R}} - {{\left( { - {{11} \over {10}}} \right)\left( { - {{11} \over {10}} - 1} \right)} \over 2}{{{a^2}} \over {{R^2}}}} \right]$$
= -4P0$$\pi $$R2$$a$$ - $${{40 \times 21} \over {100 \times 2}}{P_0}\pi R{a^2}$$
= -4P0$$\pi $$R2$$a$$ - $$4.2{P_0}\pi R{a^2}$$
Wwater = P0dV
= P0$${\left[ {{4 \over 3}\pi {R^3} - {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {{R^3} - {{\left( {R - a} \right)}^3}} \right]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {\left[ {R - \left( {R - a} \right)} \right]\left[ {{R^2} + R\left( {R - a} \right) + {{\left( {R - a} \right)}^2}} \right]} \right]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {\left[ a \right]\left[ {3{R^2} - 3Ra + {a^2}} \right]} \right]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {\left( {3{R^2}a - 3R{a^2} + {a^3}} \right)} \right]}$$
= $${4{P_0}\pi \left[ {{R^2}a - R{a^2}} \right]}$$ [ignore $${{a^3}}$$ term as no $${{a^3}}$$ term in the question]
$$ \therefore $$ Wgas + Wwater = -$$4{P_0}\pi R{a^2}$$ - $$4.2{P_0}\pi R{a^2}$$
= $$ - 4{P_0}\pi R{a^2}\left[ {1 + 1.05} \right]$$
= $$ - 4{P_0}\pi R{a^2}\left[ {2.05} \right]$$
From (i), Wexternal = - (Wgas + Wwater)
= $$4{P_0}\pi R{a^2}\left[ {2.05} \right]$$
$$ \therefore $$ X = 2.05
$$ \therefore $$ Wexternal + Wwater + Wgas = 0 ....(i)
where Wexternal = work done by external agent
Wwater = work done by water
Wgas = work done by gas
For adiabatic process, $${P_1}V_1^\gamma = {P_2}V_2^\gamma $$
Here, $${P_0}{\left[ {{4 \over 3}\pi {R^3}} \right]^{{{41} \over {30}}}} = P{\left[ {{4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]^{{{41} \over {30}}}}$$
$$ \Rightarrow $$ P = $${P_0}{\left[ {{R \over {R - a}}} \right]^{{{41} \over {10}}}}$$
Now, Wgas = $${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$$
= $${{{P_0} \times {4 \over 3}\pi {R^3} - P \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$$
= $${{{P_0} \times {4 \over 3}\pi {R^3} - {P_0}{{\left[ {{R \over {R - a}}} \right]}^{{{41} \over {10}}}} \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$$
= $${{{P_0} \times {4 \over 3}\pi {R^3}\left[ {1 - {{\left( {{R \over {R - a}}} \right)}^{{{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]} \over {{{11} \over {30}}}}$$
= $${{{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {{{R - a} \over R}} \right)}^{ - {{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]}$$
= $${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {1 - {a \over R}} \right)}^{ - {{11} \over {10}}}}} \right]$$
= $${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - 1 - {{11a} \over {10R}} - {{\left( { - {{11} \over {10}}} \right)\left( { - {{11} \over {10}} - 1} \right)} \over 2}{{{a^2}} \over {{R^2}}}} \right]$$
= -4P0$$\pi $$R2$$a$$ - $${{40 \times 21} \over {100 \times 2}}{P_0}\pi R{a^2}$$
= -4P0$$\pi $$R2$$a$$ - $$4.2{P_0}\pi R{a^2}$$
Wwater = P0dV
= P0$${\left[ {{4 \over 3}\pi {R^3} - {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {{R^3} - {{\left( {R - a} \right)}^3}} \right]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {\left[ {R - \left( {R - a} \right)} \right]\left[ {{R^2} + R\left( {R - a} \right) + {{\left( {R - a} \right)}^2}} \right]} \right]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {\left[ a \right]\left[ {3{R^2} - 3Ra + {a^2}} \right]} \right]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {\left( {3{R^2}a - 3R{a^2} + {a^3}} \right)} \right]}$$
= $${4{P_0}\pi \left[ {{R^2}a - R{a^2}} \right]}$$ [ignore $${{a^3}}$$ term as no $${{a^3}}$$ term in the question]
$$ \therefore $$ Wgas + Wwater = -$$4{P_0}\pi R{a^2}$$ - $$4.2{P_0}\pi R{a^2}$$
= $$ - 4{P_0}\pi R{a^2}\left[ {1 + 1.05} \right]$$
= $$ - 4{P_0}\pi R{a^2}\left[ {2.05} \right]$$
From (i), Wexternal = - (Wgas + Wwater)
= $$4{P_0}\pi R{a^2}\left[ {2.05} \right]$$
$$ \therefore $$ X = 2.05