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Properties of Matter question

2022 · Shift 2 · Q47
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Properties of Matter question

2022 · Shift 2 · Q47

JEE AdvancedPhysicsProperties of MatterMultiple correct+4 / −2
A bubble has surface tension SSS. The ideal gas inside the bubble has ratio of specific heats γ=53\gamma=\frac{5}{3}γ=35​. The bubble is exposed to the atmosphere and it always retains its spherical shape. When the atmospheric pressure is Pa1P_{a 1}Pa1​, the radius of the bubble is found to be r1r_{1}r1​ and the temperature of the enclosed gas is T1T_{1}T1​. When the atmospheric pressure is Pa2P_{a 2}Pa2​, the radius of the bubble and the temperature of the enclosed gas are r2r_{2}r2​ and T2T_{2}T2​, respectively. Which of the following statement(s) is(are) correct?
  1. A
    If the surface of the bubble is a perfect heat insulator, then (r1r2)5=Pa2+2Sr2Pa1+2Sr1\left(\frac{r_{1}}{r_{2}}\right)^{5}=\frac{P_{a 2}+\frac{2 S}{r_{2}}}{P_{a 1}+\frac{2 S}{r_{1}}}(r2​r1​​)5=Pa1​+r1​2S​Pa2​+r2​2S​​.
  2. B
    If the surface of the bubble is a perfect heat insulator, then the total internal energy of the bubble including its surface energy does not change with the external atmospheric pressure.
  3. C
    If the surface of the bubble is a perfect heat conductor and the change in atmospheric temperature is negligible, then (r1r2)3=Pa2+4Sr2Pa1+4Sr1\left(\frac{r_{1}}{r_{2}}\right)^{3}=\frac{P_{a 2}+\frac{4 S}{r_{2}}}{P_{a 1}+\frac{4 S}{r_{1}}}(r2​r1​​)3=Pa1​+r1​4S​Pa2​+r2​4S​​.
  4. D
    If the surface of the bubble is a perfect heat insulator, then (T2T1)52=Pa2+4Sr2Pa1+4Sr1\left(\frac{T_{2}}{T_{1}}\right)^{\frac{5}{2}}=\frac{P_{a 2}+\frac{4 S}{r_{2}}}{P_{a 1}+\frac{4 S}{r_{1}}}(T1​T2​​)25​=Pa1​+r1​4S​Pa2​+r2​4S​​.
View written solutionFree

Correct answer: C, D

  1. Pressure inside the bubble

A spherical soap bubble has two surfaces, so the excess pressure is

P in−Pa=4Sr.P_{\,in}-P_a=\frac{4S}{r}.Pin​−Pa​=r4S​.

Hence at any state,

P=Pa+4Sr,P= P_a+\frac{4S}{r},P=Pa​+r4S​,

where PPP is the pressure of the gas inside the bubble.

So for the two states,

P1=Pa1+4Sr1,P2=Pa2+4Sr2.P_1=P_{a1}+\frac{4S}{r_1},\qquad P_2=P_{a2}+\frac{4S}{r_2}.P1​=Pa1​+r1​4S​,P2​=Pa2​+r2​4S​.

Also volume of the bubble is

V=43πr3.V=\frac{4}{3}\pi r^3.V=34​πr3.
  1. Check option A: insulated bubble

If the surface is a perfect heat insulator, the process is adiabatic for the enclosed ideal gas:

PVγ=constant,γ=53.PV^\gamma=\text{constant},\qquad \gamma=\frac53.PVγ=constant,γ=35​.

Thus,

P1V15/3=P2V25/3.P_1V_1^{5/3}=P_2V_2^{5/3}.P1​V15/3​=P2​V25/3​.

Since V∝r3V\propto r^3V∝r3,

V5/3∝r5.V^{5/3}\propto r^5.V5/3∝r5.

Therefore,

P1r15=P2r25P_1r_1^5=P_2r_2^5P1​r15​=P2​r25​

which gives

(r1r2)5=P2P1=Pa2+4Sr2Pa1+4Sr1.\left(\frac{r_1}{r_2}\right)^5=\frac{P_2}{P_1} =\frac{P_{a2}+\frac{4S}{r_2}}{P_{a1}+\frac{4S}{r_1}}.(r2​r1​​)5=P1​P2​​=Pa1​+r1​4S​Pa2​+r2​4S​​.

But option A states

(r1r2)5=Pa2+2Sr2Pa1+2Sr1,\left(\frac{r_1}{r_2}\right)^5=\frac{P_{a2}+\frac{2S}{r_2}}{P_{a1}+\frac{2S}{r_1}},(r2​r1​​)5=Pa1​+r1​2S​Pa2​+r2​2S​​,

which uses 2S/r2S/r2S/r instead of 4S/r4S/r4S/r.

So A is false.


  1. Check option B: insulated bubble and total internal energy including surface energy

For a soap bubble, total relevant energy is

  • internal energy of gas: UUU,
  • surface energy: Es=2(4πr2)S=8πSr2E_s=2(4\pi r^2)S=8\pi Sr^2Es​=2(4πr2)S=8πSr2.

For an adiabatic change, δQ=0\delta Q=0δQ=0. If the bubble changes quasi-statically,

dU=−P dV,dU = -P\,dV,dU=−PdV,

where PPP is gas pressure inside.

Now consider total energy

E=U+8πSr2.E=U+8\pi Sr^2.E=U+8πSr2.

Then

dE=dU+d(8πSr2)=−P dV+16πSr dr.dE=dU+d(8\pi Sr^2)=-P\,dV+16\pi Sr\,dr.dE=dU+d(8πSr2)=−PdV+16πSrdr.

Since

dV=4πr2dr,dV=4\pi r^2dr,dV=4πr2dr,

we get

dE=−P(4πr2dr)+16πSr dr.dE= -P(4\pi r^2dr)+16\pi Sr\,dr.dE=−P(4πr2dr)+16πSrdr.

Using

P=Pa+4Sr,P=P_a+\frac{4S}{r},P=Pa​+r4S​, dE=−(Pa+4Sr)4πr2dr+16πSr drdE=-\left(P_a+\frac{4S}{r}\right)4\pi r^2dr+16\pi Sr\,drdE=−(Pa​+r4S​)4πr2dr+16πSrdr dE=−4πr2Pa dr−16πSr dr+16πSr drdE=-4\pi r^2P_a\,dr-16\pi Sr\,dr+16\pi Sr\,drdE=−4πr2Pa​dr−16πSrdr+16πSrdr dE=−Pa dV.dE=-P_a\,dV.dE=−Pa​dV.

So total energy of gas + surface energy is not constant; it changes due to work done against atmosphere.

Hence B is false.


  1. Check option C: perfectly conducting surface, atmospheric temperature unchanged

If the surface is a perfect heat conductor and atmospheric temperature change is negligible, the gas remains at constant temperature. So process is isothermal:

P1V1=P2V2.P_1V_1=P_2V_2.P1​V1​=P2​V2​.

Since V∝r3V\propto r^3V∝r3,

P1r13=P2r23.P_1r_1^3=P_2r_2^3.P1​r13​=P2​r23​.

Thus,

(r1r2)3=P2P1=Pa2+4Sr2Pa1+4Sr1.\left(\frac{r_1}{r_2}\right)^3=\frac{P_2}{P_1} =\frac{P_{a2}+\frac{4S}{r_2}}{P_{a1}+\frac{4S}{r_1}}.(r2​r1​​)3=P1​P2​​=Pa1​+r1​4S​Pa2​+r2​4S​​.

This matches option C.

So C is true.


  1. Check option D: insulated bubble relation between temperatures

For adiabatic process of an ideal gas,

TVγ−1=constant.TV^{\gamma-1}=\text{constant}.TVγ−1=constant.

With γ=5/3\gamma=5/3γ=5/3,

TV2/3=constant.TV^{2/3}=\text{constant}.TV2/3=constant.

Since V∝r3V\propto r^3V∝r3, we have

V2/3∝r2.V^{2/3}\propto r^2.V2/3∝r2.

Hence,

Tr2=constant.Tr^2=\text{constant}.Tr2=constant.

So,

T2T1=r12r22.\frac{T_2}{T_1}=\frac{r_1^2}{r_2^2}.T1​T2​​=r22​r12​​.

Therefore,

(T2T1)5/2=(r12r22)5/2=(r1r2)5.\left(\frac{T_2}{T_1}\right)^{5/2}=\left(\frac{r_1^2}{r_2^2}\right)^{5/2}=\left(\frac{r_1}{r_2}\right)^5.(T1​T2​​)5/2=(r22​r12​​)5/2=(r2​r1​​)5.

From adiabatic relation already found,

(r1r2)5=Pa2+4Sr2Pa1+4Sr1.\left(\frac{r_1}{r_2}\right)^5=\frac{P_{a2}+\frac{4S}{r_2}}{P_{a1}+\frac{4S}{r_1}}.(r2​r1​​)5=Pa1​+r1​4S​Pa2​+r2​4S​​.

Thus,

(T2T1)5/2=Pa2+4Sr2Pa1+4Sr1.\left(\frac{T_2}{T_1}\right)^{5/2}=\frac{P_{a2}+\frac{4S}{r_2}}{P_{a1}+\frac{4S}{r_1}}.(T1​T2​​)5/2=Pa1​+r1​4S​Pa2​+r2​4S​​.

This matches option D.

So D is true.


  1. Final conclusion

Correct statements are:

C,D\boxed{C, D}C,D​

These agree with the stored answer.

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