Question

At the bottom of a lake where temperature is 7°C
the pressure is 2.8 atmosphere. An air bubble of
radius 1 cm at the bottom rises to the surface
where the temperature is 27° C. Radius of air
bubble at the surface is
31/3
41/3
51/3
613​

Answer 1

Answer:

31/3

Explanation:

Answer 2

Answer:

${3}^{\dfrac{1}{3}} cm$

Explanation:

At the bottom of a lake =>

$temperature = t_1 = 7^{\circ} C = 280 K$

$pressure = p_1 = 2.8 atm$

$Radius = r_1= 1cm$

An air bubble of =>

$temperature =t_2= 27 ^{\circ} C = 300 K$

$pressure = p_2 = 1atm$

$radius = r_2 = ?$

so by the formula

$\dfrac {p_1 v_1}{t_1} = \dfrac {p_2 v_2}{t_2}$

$\dfrac{2.8 \times 4/3 \pi {1}^{3}}{280} = \dfrac{ 1\times4/3 \pi {r}^{3}}{300}$

$r_2 =( 2.8 \times \dfrac{30}{28})^\dfrac{1}{3}$

therefore,

${r_2 = 3^\dfrac{1}{3} }} cm}}$