Question 1: Thermodynamics & KTG

Question

A spherical bubble inside water has radius R. Take the pressure inside the bubble and the water pressure to be p
0
. 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$\pi$p
0
Ra
2
)X, where X is a constant and $\gamma$ = C
p
/C
v
= 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

Integer / Numeric Answer Type