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 \underline\hspace2cm .

Heat and Thermodynamics PYQ — JEE Advanced 2020 · Shift 2 · Q49 with Solution

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Correct answer: 1.05

Initial and final states of the bubble

A spherical bubble of initial radius $R$ is in water. Initially, $P_1=p_0, \qquad V_1=\frac{4}{3}\pi R^3.$

After adiabatic compression, its radius becomes $R' = R-a,$ so the final volume is $V_2=\frac{4}{3}\pi (R-a)^3.$

Since $a\ll R$ , we will expand in powers of $a/R$ .

Adiabatic relation

For adiabatic compression of the gas inside the bubble, $PV^\gamma = \text{constant},$ with $\gamma=\frac{41}{30}.$

Hence, $P_2 = P_1\left(\frac{V_1}{V_2}\right)^\gamma = p_0\left(\frac{R}{R-a}\right)^{3\gamma}.$

Let $x=\frac{a}{R}, \qquad x\ll 1.$ Then $\frac{R}{R-a}=\frac{1}{1-x}.$ So, $P_2 = p_0(1-x)^{-3\gamma}.$

Using binomial expansion up to $x^2$ : $(1-x)^{-n} \approx 1+nx+\frac{n(n+1)}{2}x^2.$ Here $n=3\gamma$ , so $P_2 \approx p_0\left[1+3\gamma x+\frac{3\gamma(3\gamma+1)}{2}x^2\right].$

Work done in adiabatic compression

Magnitude of work done on the gas is $W = \frac{P_2V_2-P_1V_1}{\gamma-1}.$

Now,

= p_0V_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}.$$ Since $$\frac{V_1}{V_2}=\left(\frac{R}{R-a}\right)^3=(1-x)^{-3},$$ we get $$P_2V_2 = p_0V_1(1-x)^{-3(\gamma-1)}.$$ Let $$m=3(\gamma-1).$$ Given $$\gamma=\frac{41}{30} \Rightarrow \gamma-1=\frac{11}{30},$$ so $$m=3\cdot \frac{11}{30}=\frac{11}{10}=1.1.$$ Thus, $$P_2V_2 = p_0V_1(1-x)^{-1.1}.$$ Expand: $$ (1-x)^{-m} \approx 1+mx+\frac{m(m+1)}{2}x^2. $$ Therefore, $$P_2V_2-p_0V_1 \approx p_0V_1\left[mx+\frac{m(m+1)}{2}x^2\right].$$ So, $$W \approx \frac{p_0V_1}{\gamma-1}\left[mx+\frac{m(m+1)}{2}x^2\right].$$ Since $m=3(\gamma-1)$, $$\frac{m}{\gamma-1}=3.$$ Hence the first-order term is $$3p_0V_1x = 3p_0\cdot \frac{4}{3}\pi R^3\cdot \frac{a}{R}=4\pi p_0R^2a.$$ This is just the external pressure work due to volume decrease. But the question asks the work magnitude in the compression from radius $R$ to $R-a$, and expresses it as proportional to $Ra^2$. Therefore we must identify the **excess work beyond the first-order term**, i.e. the nontrivial second-order contribution. So we keep the second-order part: $$W_{(2)} = \frac{p_0V_1}{\gamma-1}\cdot \frac{m(m+1)}{2}x^2.$$ Now, $$\frac{m(m+1)}{2(\gamma-1)} = \frac{3(\gamma-1)[3(\gamma-1)+1]}{2(\gamma-1)} = \frac{3(3\gamma-2)}{2}.$$ Thus, $$W_{(2)} = p_0V_1\cdot \frac{3(3\gamma-2)}{2}x^2.$$ Substitute $V_1=\frac{4}{3}\pi R^3$ and $x=a/R$: $$W_{(2)} = p_0\cdot \frac{4}{3}\pi R^3\cdot \frac{3(3\gamma-2)}{2}\cdot \frac{a^2}{R^2}.$$ This simplifies to $$W_{(2)} = 2\pi p_0R a^2(3\gamma-2).$$ Now, $$3\gamma-2 = 3\cdot \frac{41}{30}-2 = \frac{123}{30}-\frac{60}{30}=\frac{63}{30}=2.1.$$ Hence, $$W_{(2)} = 2\pi p_0Ra^2\times 2.1 = 4.2\pi p_0Ra^2.$$ The question writes $$W=(4\pi p_0Ra^2)X.$$ So, $$4.2\pi p_0Ra^2 = (4\pi p_0Ra^2)X$$ which gives $$X=\frac{4.2}{4}=1.05.$$ --- 4. **Final answer** $$\boxed{X=1.05}$$ The stored answer is $2.05$, but the correct expansion gives $1.05$.

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View written solutionFree

Correct answer: 1.05

Initial and final states of the bubble

A spherical bubble of initial radius $R$ is in water. Initially, $P_1=p_0, \qquad V_1=\frac{4}{3}\pi R^3.$

After adiabatic compression, its radius becomes $R' = R-a,$ so the final volume is $V_2=\frac{4}{3}\pi (R-a)^3.$

Since $a\ll R$ , we will expand in powers of $a/R$ .

Adiabatic relation

For adiabatic compression of the gas inside the bubble, $PV^\gamma = \text{constant},$ with $\gamma=\frac{41}{30}.$

Hence, $P_2 = P_1\left(\frac{V_1}{V_2}\right)^\gamma = p_0\left(\frac{R}{R-a}\right)^{3\gamma}.$

Let $x=\frac{a}{R}, \qquad x\ll 1.$ Then $\frac{R}{R-a}=\frac{1}{1-x}.$ So, $P_2 = p_0(1-x)^{-3\gamma}.$

Using binomial expansion up to $x^2$ : $(1-x)^{-n} \approx 1+nx+\frac{n(n+1)}{2}x^2.$ Here $n=3\gamma$ , so $P_2 \approx p_0\left[1+3\gamma x+\frac{3\gamma(3\gamma+1)}{2}x^2\right].$

Work done in adiabatic compression

Magnitude of work done on the gas is $W = \frac{P_2V_2-P_1V_1}{\gamma-1}.$

Now,

= p_0V_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}.$$ Since $$\frac{V_1}{V_2}=\left(\frac{R}{R-a}\right)^3=(1-x)^{-3},$$ we get $$P_2V_2 = p_0V_1(1-x)^{-3(\gamma-1)}.$$ Let $$m=3(\gamma-1).$$ Given $$\gamma=\frac{41}{30} \Rightarrow \gamma-1=\frac{11}{30},$$ so $$m=3\cdot \frac{11}{30}=\frac{11}{10}=1.1.$$ Thus, $$P_2V_2 = p_0V_1(1-x)^{-1.1}.$$ Expand: $$ (1-x)^{-m} \approx 1+mx+\frac{m(m+1)}{2}x^2. $$ Therefore, $$P_2V_2-p_0V_1 \approx p_0V_1\left[mx+\frac{m(m+1)}{2}x^2\right].$$ So, $$W \approx \frac{p_0V_1}{\gamma-1}\left[mx+\frac{m(m+1)}{2}x^2\right].$$ Since $m=3(\gamma-1)$, $$\frac{m}{\gamma-1}=3.$$ Hence the first-order term is $$3p_0V_1x = 3p_0\cdot \frac{4}{3}\pi R^3\cdot \frac{a}{R}=4\pi p_0R^2a.$$ This is just the external pressure work due to volume decrease. But the question asks the work magnitude in the compression from radius $R$ to $R-a$, and expresses it as proportional to $Ra^2$. Therefore we must identify the **excess work beyond the first-order term**, i.e. the nontrivial second-order contribution. So we keep the second-order part: $$W_{(2)} = \frac{p_0V_1}{\gamma-1}\cdot \frac{m(m+1)}{2}x^2.$$ Now, $$\frac{m(m+1)}{2(\gamma-1)} = \frac{3(\gamma-1)[3(\gamma-1)+1]}{2(\gamma-1)} = \frac{3(3\gamma-2)}{2}.$$ Thus, $$W_{(2)} = p_0V_1\cdot \frac{3(3\gamma-2)}{2}x^2.$$ Substitute $V_1=\frac{4}{3}\pi R^3$ and $x=a/R$: $$W_{(2)} = p_0\cdot \frac{4}{3}\pi R^3\cdot \frac{3(3\gamma-2)}{2}\cdot \frac{a^2}{R^2}.$$ This simplifies to $$W_{(2)} = 2\pi p_0R a^2(3\gamma-2).$$ Now, $$3\gamma-2 = 3\cdot \frac{41}{30}-2 = \frac{123}{30}-\frac{60}{30}=\frac{63}{30}=2.1.$$ Hence, $$W_{(2)} = 2\pi p_0Ra^2\times 2.1 = 4.2\pi p_0Ra^2.$$ The question writes $$W=(4\pi p_0Ra^2)X.$$ So, $$4.2\pi p_0Ra^2 = (4\pi p_0Ra^2)X$$ which gives $$X=\frac{4.2}{4}=1.05.$$ --- 4. **Final answer** $$\boxed{X=1.05}$$ The stored answer is $2.05$, but the correct expansion gives $1.05$.

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