Ask question

Ask question

All categories

3 years ago

7

If you use such a tank to fill 0.020 m3 foil balloons (which don't stretch, and so have an internal pressure that is very close

to atmospheric pressure), how many balloons can you expect to fill? Assume the temperature is 20∘C.

1 answer:

Arte-miy333 [17]3 years ago

8 0

The question is incomplete, here is a complete question.

Party stores sell small tanks containing 41 g of helium gas.

If you use such a tank to fill 0.020 m³ foil balloons (which don't stretch, and so have an internal pressure that is very close to atmospheric pressure), how many balloons can you expect to fill? Assume the temperature is 20°C.

Answer : The number of balloon filled can be, 12 balloons.

Explanation :

First we have to calculate the number of moles of gas in balloon.

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm (atmospheric pressure)

V = volume of gas = $0.020m^3=20.0L$ $(1m^3=1000L)$

T = temperature of gas = $20^oC=273+20=293K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (20.0L)=n\times (0.0821L.atm/mol.K)\times (293K)$

$n=0.831mol$

Thus, the number of moles of gas in balloon is 0.831 mol per balloon.

Now we have to calculate the mass of gas in balloon.

$\text{Mass of gas}=\text{Moles of gas}\times \text{Molar mass of gas}$

Molar mass of He gas = 4 g/mole

$\text{Mass of gas}=(0.831 mol/balloon)\times 4g/mol=3.32g/balloon$

Now we have to calculate the number of balloon can be filled.

Number of balloon filled = $\frac{41g}{3.32g/balloon}=12.3balloon\approx 12balloons$

Therefore, the number of balloon filled can be, 12 balloons.

You might be interested in

Which of the following statements is true for real gases?

jok3333 [9.3K]

Answer:

A. The volume occupied by the molecules can cause an increase in pressure compared to the ideal gas.

D. As attractive forces between molecules increase, deviations from ideal behavior become more apparent at relatively low temperatures.

Explanation:

as we know by real gas equation

$(P + \frac{an^2}{V^2})(V - nb) = nRT$

while ideal gas equation is given as

$PV = nRT$

so from above formula we can say that net pressure is increased in real gas

Also we know that all real gas will close to behave like ideal gas when the pressure of the real gas is low and temperature of the gas is high

So above are the correct observations

3 0

2 years ago

Read 2 more answers

Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds.

s344n2d4d5 [400]

This question is incomplete, the complete question is;

A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t² + 8√t, where t is the time in seconds and s(t) is the distance in feet from the top of the building.

(a) Write a formula for the average velocity of the ball for t near 4.

(b) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds

(c) What is your estimate for the instantaneous velocity of the ball at t = 4

Answer:

a)

Average velocity, (Vavg) of the ball for t near 4.

Vavg = [s(4) - s(0)] / (4 - 0)

Where s(4) = -16 × 4² + 8 × √4= - 240 m

s(0) = -16 × 0 + 8 * 0 = 0

b)

duration = 1 sec

Vavg = [s(5) - s(4)] / (5 - 4)

s(5) = -16 × 52 + 8 × √5 = - 382 m

s(4) = -16 × 42 + 8 √4 = - 240 m

Vavg = (-382 - (-240)) / (5 - 4)

Vavg = - 142.1 m/s

duration = 0.5 sec

Vavg = [s(4.5) - s(4)] / (4.5 - 4)

s(4.5) = -16 × 4.52 + 8 × √4.5 = - 307 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-307 - (-240)) / (4.5 - 4)

Vavg= - 134.1 m/s

duration = 0.05 sec

Vavg = [s(4.05) - s(4)] / (4.05 - 4)

s(4.05) = -16 × 4.052 + 8 × √4.05 = - 246 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-246 - (-240)) / (4.05 - 4)

Vavg= - 126.8 m/s

c)

Instantaneous velocity, v = ds/dt

= - 16 × 2 × t + 8 ×× (0.5 / √t )

= - 32 × t + 4/√t

ds/dt at t = 4 is,

v = - 32 × 4 + 4 / √4

= - 126 m/s

5 0

3 years ago

Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals,

Andre45 [30]

Answer:

9241.6 W or 12.39318 hp

Explanation:

u = Initial velocity = 0

v = Final velocity

m = Mass

t = Time taken

Energy

$KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules$

Power

$P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W$

Converting to hp

$1\ W=\frac{1}{745.7}\ hp$

$\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp$

The power developed by the cheetah is 9241.6 W or 12.39318 hp

7 0

3 years ago

Speakers A and B are vibrating in phase. They are directly facing each other, are 8.2 m apart, and are each playing a 78.0 Hz to

Stels [109]

Answer:6.298,4.1,1.9015

Explanation:

Wavelength= $\frac{velocity of sound }{frequency}$

= $\frac{343}{78}$ =4.397 m

Distance of 3rd speaker from speaker A is x

From B 78-x

Difference between the distances must be a whole number of wavelengths

First

$x-\left ( 8.2-x\right )=4.397$ for 1 st wavelength

2x=8.2+4.397=12.597

x=6.298m

second

For zero wavelength

$x-\left ( 8.2-x\right )=0$

2x=8.2

x=4.1m

Third

$\left ( 8.2-x\right )-x=4.397$

x=1.9015 m

6 0

2 years ago

A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w

nirvana33 [79]

Answer:

$|D_{depth} |=19.697m$

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

$x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\ -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s$

As we have find the time taken now we need to find the final velocity vf from below equation as

$v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s$

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

$|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m$

6 0

3 years ago