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2 years ago

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At 4.00 l, an expandable vessel contains 0.864 mol of oxygen gas. how many liters of oxygen gas must be added at constant temper

ature and pressure if you need a total of 1.60 mol of oxygen gas in the vessel

1 answer:

svet-max [94.6K]2 years ago

6 0

Givens
=====
V = 4.00 L
T = 273oK We're assuming the temperature does not change, just the pressure.
n = 0.864 moles
R = 8.314 joules / mole * oK
P = ?????

Formula
======
PV = n*R*T
P = n*R*T/V
P = 0.864 * 8.314 * 273 / 4
P = 490 kpa

You have to add 1.6 – 0.864 = 0.736 moles of gas.

We have to assume that the temperature and pressure remain the same when we add the 0.736 moles of gas. We are now looking for the volume.

PV = n*R*T

<span> V = 0.736 * 8.314 * 273 / 490</span>

V = 3.41 L Remember this is at about 4 atmospheres so we have to convert to Standard Pressure.

Total Volume = 3.41 + 4.00 = 4.41

V1 * P1 = V2 * P2

P1 = 490 kPa

P2 = 101 kPa

V1 = 7.41 L

V2 = ????

<span> <span> 7.41* 490 = V2 * 101 V2 = 7.41 * 490 / 101 V2 = 35.94 L </span> </span>

<span>You had 4 L now you need 31.94 more.</span>

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Answer:

The increase in the gravitational potential energy is 29.93 joules.

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Given that,

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A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

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$\mu =0.75$

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Equating both forces

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Simplifying

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3 years ago

He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How

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Answer:

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A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

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e) 1.92 m/sec

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The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

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Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 = 9.8t^2 / 2

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3 = 7.67t therefore t = 0.39 sec

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Answer:

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