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3 years ago

3.27 moles of an ideal gas in a 50.0 L tank has a pressure of 171000 Pa. What is the temperature of the gas? (Unit=degrees C)

Physics

1 answer:

laiz [17]3 years ago

3 0

The temperature of the gas is 41.3 °C.

Answer:

The temperature of the gas is 41.3 °C.

Explanation:

So on combining the Boyle's and Charles law, we get the ideal law of gas that is PV=nRT. Here P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature. The SI unit of pressure is atm. So we need to convert 1 Pa to 1 atm, that is 1 Pa = 9.86923× $10^{-6}$ atm. Thus, 171000 Pa = 1.6876 atm.

We know that the gas constant R = 0.0821 atmLMol–¹K-¹. Then the volume of the gas is given as 50 L and moles are given as 3.27 moles.

Then substituting all the values in ideal gas equation ,we get

1.6876×50=3.27×0.0821×T

Temperature = $\frac{84.38}{0.268467} =314.3 K$

So the temperature is obtained to be 314.3 K. As 0°C = 273 K,

Then 314.3 K = 314.3-273 °C=41.3 °C.

Thus, the temperature is 41.3 °C.

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Lerok [7]

Answer:

Option A

Explanation:

The Equation represents the displacement of the object which is represented by x

$x=t-t^2$

so, $x_0$ means when time is zero so we replace t with zero in the equation,

$x_0=(0)-(0)^2\\x_0=0$

now for v which is velocity we need to differentiate the function as the formula for velocity is rate of change of displacement over time so we derivate the equation once and get,

$v=1-2t\\$

now for $v_0$ we insert t = 0 and get

$v_0=1-2(0)\\v_0=1$

now for a which is acceleration the formula of acceleration is rate of change of velocity over time, so we differentiate the the equation of v(velocity) once or the equation of x(displacement) twice so now we get,

$a=-2$

so Option A is your answer.

Remember derivative of a constant is always zero because a constant value has no rate of change has its a constant hence the derivative is 0

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3 years ago

A.stationary <br> B. Accelerating <br> C. Decelerating <br> D. Moving at constant speed

V125BC [204]

Answer:

ACCELERATING OR DECELERATING

Explanation:

I'M NOT SURE

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3 years ago

A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc

katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

$t=\frac{5.60km}{4.85km/hr}=1.155hr$

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

$V_{ds}=V_{river}+V{boat}$

$V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr$

and now up stream:

$V_{us}=V_{boat}-V{river}$

$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

$t_{ds}=\frac{x}{v_{ds}}$

$t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr$

and the time up stream:

$t_{us}=\frac{x}{v_{us}}$

$t_{us}=\frac{2.80km}{2,80km/hr}=1hr$

so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$

$t=\frac{5.60km}{5.60km/hr}$

t= 1hr

4 0

3 years ago

Read 2 more answers

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

The horizontal component of the pulling force, $F cos \theta$ , where F = 50 N is the magnitude and $\theta=60^{\circ}$ is the angle between the direction of the force and the horizontal; this force acts in the forward direction
The force of friction, $F_f$ , acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$