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Vinvika [58]
3 years ago
9

An ideal gas is confined within a closed cylinder at atmospheric pressure (1.013 * 105 Pa) by a piston. The piston moves until t

he volume of the gas is reduced to one-ninth of the initial volume. What is the final pressure of the gas when its temperature returns to its initial value?
Physics
1 answer:
likoan [24]3 years ago
5 0

Answer:

911700\ \text{Pa}

Explanation:

P_1 = Initial pressure = 1.013\times 10^5\ \text{Pa}

V_1 = Initial volume

V_2= Final volume = \dfrac{V_1}{9}\\\Rightarrow \dfrac{V_1}{V_2}=9

Temperature is the same in the initial and final state

From the ideal gas law we have

P_1V_1=P_2V_2\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=P_1\times9\\\Rightarrow P_2=1.013\times 10^5\times 9\\\Rightarrow P_2=911700\ \text{Pa}

The final pressure of the system is 911700\ \text{Pa}.

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Answer:

1.  F = M x A

2. Force  

3. 2nd Law: Force

4. a, b, c (in order)

5. 3rd Law: Action and Reaction

6. b, c, a (in order)

7. 1st Law: Inertia

3 0
2 years ago
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A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric fiel
nexus9112 [7]

Answer:

The electric flux is 280\ \rm N.m^2/C

Explanation:

Given:

  • Radius of the disc R=0.50 m
  • Angle made by disk with the horizontal \theta=30^\circ
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The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows

\phi=\int E.dA

where

  • \phi is the total Electric Flux
  • E is the Electric Field
  • dA is the Area through which the electric flux is to be calculated.

Now according to question we have

=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C

Hence the electric flux is calculated.

8 0
3 years ago
A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres
Lemur [1.5K]

Answer:

V_2 = 0.125 m^3

Work done =  = 5 kJ

Explanation:

Given data:

volume of nitrogen v_1 = 0.08 m^3

P_1 = 150 kPa

T_1 = 200 degree celcius = 473 Kelvin

P_2 = 80 kPa

Polytropic exponent n = 1.4

\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}

putting all value

\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}

\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS

polytropic process is given as

P_1 V_1^n = P_2 V_2^n

150\times 0.08^{1.4} = 80 \times V_2^{1.4}

V_2 = 0.125 m^3

work done = \frac{P_1 V_1 -P_2 V_2}{n-1}

= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}

                  = 5 kJ

4 0
3 years ago
An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. Howeve
Vlad1618 [11]

Answer:

v=0.9833\ c

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The density changes means that the length in the direction of the motion is changed.

Therefore,

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Given :

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Mass, m = 3.3 kg

Density = 8100 kg/m^3

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$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$

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$0.033=  1-\frac{v^2}{c^2}}$

$\frac{v^2}{c^2}= 0.967$

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v=0.9833\ c

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Answer:

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