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3 years ago

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Consider two samples of gas. Sample A consists of n moles and is kept at temperature T in container of volume V. Gas B consists

of 2n moles, and is kept under the temperature T/2 in a container of volume V/2. What is the pressure of gas B in terms of pressure of gas A, P?
a. 2P
b. 4P
c. 8P
d. P
e. P/2
f. P/4
g. P/8

1 answer:

MrRissso [65]3 years ago

6 0

Answer: e. P/2

Explanation:

For ideal gases, we have the relation:

P*V = n*R*T

where:

n = number of mols

R = Gas constant

T = temperature

V = volume

P = pressure.

We know that for sample A, we have n moles, a temperature T and a volume V, then the pressure of this sample will be:

Pa = (n*R*T)/V.

For sample B, we have:

n/2 moles, temperature T/2 and a volume V/2, then the pressure will be:

Pb = (n/2)*R*(T/2)*(2/V) = (n*R*T/V)*(2/4)

and:

(n*R*T/V) = Pa

Then we can replace it and we get:

Pb = (n*R*T/V)*(2/4) = Pa*(2/4) = Pa*(1/2) = Pa/2.

Then the correct option is e.

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3. A football is kicked with a speed of 35 m/s at an angle of 40°.

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a) 22.5 m/s

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$u_y = u sin \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_y = (35)(sin 40)=22.5 m/s$

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The initial horizontal velocity is given by:

$u_x = u cos \theta$

where

u = 35 m/s is the initial speed

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Substituting into the equation, we find

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The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

$v_y = u_y + at$

where

$v_y$ is the vertical velocity at time t

$u_y = 22.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, $v_y =0$ ; substituting, we find the time t at which this happens:

$0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s$

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s is the vertical displacement at time t

$u_y = 22.5 m/s$

$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

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First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

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Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

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Therefore, the horizontal distance travelled during the whole motion is

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So, the ball lands 123.3 m far from the initial point.

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