An air bubble of volume 2.9 cm^3 rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17^ C. The volume of the bubble when it reaches the surface,…

Given: - Initial volume at the bottom, V1 = 2.9 cm^3 - Depth of pool, h = 5 m - Bottom temperature, T1 = 17 + 273 = 290 K - Surface temperature, T2 = 27 + 273 = 300 K Find: Final volume V2 at the surface. At the bottom, P1 = P0 + g h = 10^5 + 10^3(10)(5) = 1.5 10^5 Pa At the surface, P2 = 10^5 Pa Using the ideal gas relation, P1 V1T1 = P2 V2T2 So, V2 = V1 P1P2 T2T1 Substituting, V2 = 2.9 ( 1.51) ( 300290) V2 = 2.9 1.5 3029 = 1.5 3 = 4.5 cm^3 Therefore, the volume of the bubble at the surface is 4.5 cm^3. The correct option is C.

An air bubble of volume 2.9 cm^3 rises from the bottom of a swimm… - Solution | JEE Physics

MCQEasyJEE 2026Equation of State of Ideal Gas

An air bubble of volume $2.9 \, \text{cm}^3$ rises from the bottom of a swimming pool of $5 \, \text{m}$ deep. At the bottom of the pool water temperature is $17^\circ \text{C}$ . The volume of the bubble when it reaches the surface, where the water temperature is $27^\circ \text{C}$ , is ___ $\text{cm}^3$ .__

A
$2.0$
B
$3.0$
C
$4.5$
D
$4.2$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

Initial volume at the bottom, $V_1 = 2.9 \, \text{cm}^3$
Depth of pool, $h = 5 \, \text{m}$
Bottom temperature, $T_1 = 17 + 273 = 290 \, \text{K}$
Surface temperature, $T_2 = 27 + 273 = 300 \, \text{K}$

Find: Final volume $V_2$ at the surface.

At the bottom,

P_1 = P_0 + \rho g h = 10^5 + 10^3(10)(5) = 1.5 \times 10^5 \, \text{Pa}

At the surface,

P_2 = 10^5 \, \text{Pa}

Using the ideal gas relation,

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

So,

V_2 = V_1 \frac{P_1}{P_2} \frac{T_2}{T_1}

Substituting,

V_2 = 2.9 \left(\frac{1.5}{1}\right) \left(\frac{300}{290}\right)

V_2 = 2.9 \times 1.5 \times \frac{30}{29} = 1.5 \times 3 = 4.5 \, \text{cm}^3

Therefore, the volume of the bubble at the surface is $4.5 \, \text{cm}^3$ . The correct option is C.

Common mistakes

Using temperature in degree Celsius directly is incorrect because gas-law relations require absolute temperature. Convert $17^\circ \text{C}$ and $27^\circ \text{C}$ to Kelvin before substitution.
Taking the bottom pressure as only atmospheric pressure is wrong because the bubble is $5 \, \text{m}$ below the surface. Use $P_1 = P_{atm} + \rho g h$ , not just $P_{atm}$ .
Assuming the process is isothermal is incorrect because the water temperature changes from $17^\circ \text{C}$ to $27^\circ \text{C}$ . Include both pressure and temperature changes through $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ .

Practice more Equation of State of Ideal Gas questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Start free trial More on Equation of State of Ideal Gas

Answer

Step-by-step solution

Common mistakes

Practice more Equation of State of Ideal Gas questions

Related questions