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Alborosie
3 years ago
7

If 41.5 mol of an ideal gas occupies 86.5 L at 27.00 °C, what is the pressure of the gas?

Physics
2 answers:
Dmitry [639]3 years ago
7 0
PV=nRT
(P)(86.5)=(41.5)(.08206)(300.15)
(P)(86.5)=(1022.157824)
P=11.81685345 atm
AlexFokin [52]3 years ago
5 0

YUP!!and I think its important to note that when using this value of R

  • <em>It is crucial to match your units of Pressure, Volume, number of mole, and Temperature with the units of R. If you use the first value of </em>R,<em> which is </em>0.082057 L <u><em>your unit for Pressure must be atm, Volume must be Liter, & Temperature must be Kelvin.</em></u>

<u><em /></u>

Clarity in Chemistry is crucial.

*_^

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a) 4 forces

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c) 246 N

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a)

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2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force (F_A): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag (R): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

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F_{net} is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

a=3.1 m/s^2 is its acceleration

Substituting, we find the net force on the bicyclist:

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c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

F_{net}=F_a-R

where:

F_{net} is the net force

F_a is the applied force (forward)

R is the air drag (backward)

In this problem we have:

F_{net}=186 N is the net force (found in part b)

R=60 N is the magnitude of the air drag

Solving for F_a, we find the force produced by the bicyclist while pedaling:

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