All categories

3 years ago

Predict using Boyle's law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa.

Physics

1 answer:

kobusy [5.1K]3 years ago

6 0

The volume of the balloon will halve

Explanation:

Boyle's law states that for an ideal gas kept at constant temperature, the pressure of the gas is proportional to its volume. Mathematically,

$pV=const.$

where

p is the gas pressure

V is the volume

The equation can also be rewritten as

$p_1 V_1 = p_2 V_2$

And if we apply it to the gas inside the balloon in this problem (assuming its temperature is constant), we have:

$p_1 = 101 kPa$ is the initial pressure at sea level (the atmospheric pressure)

$V_1$ is the initial volume

$p_2 = 202 kPa$ is the final pressure

$V_2$ is the final volume

Substituting into the equation, we find:

$V_2 = \frac{p_1 V_1}{p_2}=\frac{(101)V_1}{202}=\frac{V_1}{2}$

Which means that the volume of the balloon will halve.

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

You might be interested in

A satellite is orbiting Earth with a distance R = 2REarth from Earth's center. If the satellite is moved to a distance R = 4REar

Maslowich

Answer:

Half

Explanation:

Given that:

radial distance of satellite from the earth, $R=2R_E$

Now, if the satellite is moved to a distance $R=4R_E$

We have the mathematical expression for the potential energy fue to gravitational field as:

$U=\frac{G.M.m}{R}$ ...................(1)

where:

$G = 6.67\times 10^{-11}\ m^3.kg^{-1}.s^{2}$

M = mass of earth

m = mass of satellite

R = radial distance of satellite

Now from eq. (1) initially we have:

$U=\frac{G.M.m}{2R_E}$

after the satellite is moved, we have:

$U'=\frac{G.M.m}{4R_E}$

$\Rightarrow U'=\frac{G.M.m}{2(2R_E)}$

$\Rightarrow U'=\frac{1}{2} \times U$

which is half of the initial condition.

3 0

3 years ago

In the reaction _S+302 +2SO3, what coefficient should be placed in front of the S to balance the reaction?

lys-0071 [83]

Answer:

B. 2

Explanation:

The reaction expression is given as:

_S + 3O₂ → 2SO₃

Now let us balanced the expression;

On the product side we have 2 moles of S

On the reactant side we should have 2moles of S

So, we put the coefficient 2 to balance the expression;

We have 6 moles oxygen on both sides

5 0

2 years ago

is set into oscillatory motion with an amplitude of 23.195 cm on a spring with a spring constant of 15.2676 N/m. The mass of the

fredd [130]

Answer:

maximum speed of the bananas is 18.8183 m/s

Explanation:

Given data

amplitude A = 23.195 cm

spring constant K = 15.2676 N/m

mass of the bananas m = 56.9816 kg

to find out

maximum speed of the bananas

solution

we know that radial oscillation frequency formula that is = √(K/A)

radial oscillation frequency = √(15.2676/23.195)

radial oscillation frequency is 0.8113125 rad/s

so maximum speed of the bananas = radial oscillation frequency × amplitude

maximum speed of the bananas = 0.8113125 × 23.195

maximum speed of the bananas is 18.8183 m/s

8 0

2 years ago

In order for a length of copper wire or a loop of copper wire to have a magnetic field, what must they have flowing through them

Nataliya [291]

Electricity.
When electricity flows through wire (such as a battery circuit) it creates a magnetic field around the wire.

6 0

3 years ago

I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to twice its length an

Nikitich [7]

Answer:

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Explanation:

Fundamental frequency = wave velocity/2L

where;

L is the length of the stretched rubber

Wave velocity = $\sqrt{\frac{T}{\frac{M}{L}}}$

Frequency (F₁) = $\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}$

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ = $\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1$

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).

7 0

2 years ago