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Natasha_Volkova [10]
3 years ago
15

0.16 mol of argon gas is admitted to an evacuated 70 cm^3 container at 30°C. The gas then undergoes an isothermal expansion to a

volume of 400 cm^3 .What is the final pressure of the gas?
Physics
1 answer:
Semmy [17]3 years ago
3 0

Answer:

The final pressure of the gas is 9.94 atm.

Explanation:

Given that,

Weight of argon = 0.16 mol

Initial volume = 70 cm³

Angle = 30°C

Final volume = 400 cm³

We need to calculate the initial pressure of gas

Using equation of ideal gas

PV=nRT

P_{i}=\dfrac{nRT}{V}

Where, P = pressure

R = gas constant

T = temperature

Put the value in the equation

P_{i}=\dfrac{0.16\times8.314\times(30+273)}{70\times10^{-6}}

P_{i}=5.75\times10^{6}\ Pa

P_{i}=56.827\ atm

We need to calculate the final temperature

Using relation pressure and volume

P_{2}=\dfrac{P_{1}V_{1}}{V_{2}}

P_{2}=\dfrac{56.827\times70}{400}

P_{2}=9.94\ atm

Hence, The final pressure of the gas is 9.94 atm.

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The net displacement at a point on the string where the pulses cross is 0.2 m.

The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.

A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.

At the post, the pulses are reflected and return along the string without losing any of their amplitude.

Now, let's say the ends are free.

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Now, since A = 0.1 m

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The correct option is (c).

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An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f
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- the weight of the plane: W=mg, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
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(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
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Answer:

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c = 3 x 10⁸ m/s

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