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MCQEasyJEE 2023Equation of State of Ideal Gas

JEE Physics 2023 Question with Solution

An air bubble of volume 1 cm31\,cm^3 rises from the bottom of a lake 40 m40\,m deep to the surface at a temperature of 12∘C12^\circ C. The atmospheric pressure is 1×105 Pa1 \times 10^5\,Pa, the density of water is 1000 kg/m31000\,kg/m^3, and g=10 m/s2g = 10\,m/s^2. The volume of the air bubble when it reaches the surface will be:

  • A

    3 cm33\,cm^3

  • B

    4 cm34\,cm^3

  • C

    2 cm32\,cm^3

  • D

    5 cm35\,cm^3

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Initial volume of the bubble at the bottom is 1 cm31\,\text{cm}^3, depth of lake is 40 m40\,\text{m}, atmospheric pressure is 1×105 Pa1 \times 10^5\,\text{Pa}, density of water is 1000 kg/m31000\,\text{kg/m}^3, and g=10 m/s2g = 10\,\text{m/s}^2.

Find: The volume of the air bubble at the surface.

At the surface, pressure is

P1=105 PaP_1 = 10^5 \, \text{Pa}

At depth, pressure is

P2=P1+ρgh=105+(1000)(10)(40)=5×105 PaP_2 = P_1 + \rho gh = 10^5 + (1000)(10)(40) = 5 \times 10^5 \, \text{Pa}

Since temperature is constant, apply Boyle's law:

P1V1=P2V2P_1V_1 = P_2V_2

Using the values as stated in the solution:

105×V2=5×105×110^5 \times V_2 = 5 \times 10^5 \times 1

So,

V2=5 cm3V_2 = 5 \, \text{cm}^3

Therefore, the volume of the air bubble when it reaches the surface is 5 cm35\,\text{cm}^3. The correct option is D.

Pressure Comparison Approach

Given: Surface pressure is 105 Pa10^5\,\text{Pa} and pressure at the bottom is greater by the hydrostatic term ρgh\rho gh.

Find: Bubble volume at the surface.

First compare the two pressures:

ρgh=1000×10×40=4×105 Pa\rho gh = 1000 \times 10 \times 40 = 4 \times 10^5 \, \text{Pa}

Hence bottom pressure is

Pbottom=105+4×105=5×105 PaP_{\text{bottom}} = 10^5 + 4 \times 10^5 = 5 \times 10^5 \, \text{Pa}

For an isothermal process, pressure and volume are inversely proportional:

PV=constantPV = \text{constant}

So the volume at lower pressure becomes larger by the same factor by which pressure decreases:

VsurfaceVbottom=PbottomPsurface=5×105105=5\frac{V_{\text{surface}}}{V_{\text{bottom}}} = \frac{P_{\text{bottom}}}{P_{\text{surface}}} = \frac{5 \times 10^5}{10^5} = 5

Thus,

Vsurface=5×1=5 cm3V_{\text{surface}} = 5 \times 1 = 5 \, \text{cm}^3

Therefore, the bubble volume at the surface is 5 cm35\,\text{cm}^3.

Common mistakes

  • Using only atmospheric pressure for the bubble at the bottom is incorrect because the water column adds hydrostatic pressure. Always use P=Patm+ρghP = P_{\text{atm}} + \rho gh at depth.

  • Reversing Boyle's law ratio gives the wrong volume. When pressure decreases as the bubble rises, volume must increase, so use the larger bottom pressure with the smaller bottom volume.

  • Confusing the positions of initial and final states leads to sign and ratio errors. Clearly label bottom and surface states before substituting into PV=constantPV = \text{constant}.

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