Question 113: Thermodynamics & KTG

Question

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 ________.

Accepted Answer

As change in the kinetic energy of the surface of the bubble is zero.
$$ 	herefore $$ W
external
+ W
water
+ W
gas
= 0 ....(i)
where W
external
= work done by external agent
W
water
= work done by water
W
gas
= work done by gas
For adiabatic process, $${P_1}V_1^\gamma  = {P_2}V_2^\gamma $$
Here, $${P_0}{\left[ {{4 \over 3}\pi {R^3}} ight]^{{{41} \over {30}}}} = P{\left[ {{4 \over 3}\pi {{\left( {R - a} ight)}^3}} ight]^{{{41} \over {30}}}}$$
$$ \Rightarrow $$ P = $${P_0}{\left[ {{R \over {R - a}}} ight]^{{{41} \over {10}}}}$$
Now, W
gas
= $${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma  - 1}}$$
= $${{{P_0} 	imes {4 \over 3}\pi {R^3} - P 	imes {4 \over 3}\pi {{\left( {R - a} ight)}^3}} \over {{{41} \over {30}} - 1}}$$
= $${{{P_0} 	imes {4 \over 3}\pi {R^3} - {P_0}{{\left[ {{R \over {R - a}}} ight]}^{{{41} \over {10}}}} 	imes {4 \over 3}\pi {{\left( {R - a} ight)}^3}} \over {{{41} \over {30}} - 1}}$$
= $${{{P_0} 	imes {4 \over 3}\pi {R^3}\left[ {1 - {{\left( {{R \over {R - a}}} ight)}^{{{41} \over {10}}}}{{\left( {{{R - a} \over R}} ight)}^3}} ight]} \over {{{11} \over {30}}}}$$
= $${{{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {{{R - a} \over R}} ight)}^{ - {{41} \over {10}}}}{{\left( {{{R - a} \over R}} ight)}^3}} ight]}$$
= $${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {1 - {a \over R}} ight)}^{ - {{11} \over {10}}}}} ight]$$
= $${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - 1 - {{11a} \over {10R}} - {{\left( { - {{11} \over {10}}} ight)\left( { - {{11} \over {10}} - 1} ight)} \over 2}{{{a^2}} \over {{R^2}}}} ight]$$
= -4P
0
$$\pi $$R
2
$$a$$ - $${{40 	imes 21} \over {100 	imes 2}}{P_0}\pi R{a^2}$$
= -4P
0
$$\pi $$R
2
$$a$$ - $$4.2{P_0}\pi R{a^2}$$
W
water
= P
0
dV
= P
0
$${\left[ {{4 \over 3}\pi {R^3} - {4 \over 3}\pi {{\left( {R - a} ight)}^3}} ight]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {{R^3} - {{\left( {R - a} ight)}^3}} ight]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {\left[ {R - \left( {R - a} ight)} ight]\left[ {{R^2} + R\left( {R - a} ight) + {{\left( {R - a} ight)}^2}} ight]} ight]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {\left[ a ight]\left[ {3{R^2} - 3Ra + {a^2}} ight]} ight]}$$
= $${{{4{P_0}} \over 3}\pi \left[ {\left( {3{R^2}a - 3R{a^2} + {a^3}} ight)} ight]}$$
= $${4{P_0}\pi \left[ {{R^2}a - R{a^2}} ight]}$$ [ignore $${{a^3}}$$ term as no $${{a^3}}$$ term in the question]
$$ 	herefore $$ W
gas
+ W
water
= -$$4{P_0}\pi R{a^2}$$ - $$4.2{P_0}\pi R{a^2}$$
= $$ - 4{P_0}\pi R{a^2}\left[ {1 + 1.05} ight]$$
= $$ - 4{P_0}\pi R{a^2}\left[ {2.05} ight]$$
From (i), W
external
= - (W
gas
+ W
water
)
= $$4{P_0}\pi R{a^2}\left[ {2.05} ight]$$
$$ 	herefore $$ X = 2.05

Thermodynamics & KTG

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