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grin007 [14]
3 years ago
10

An ideal monatomic gas expands from an initial pressure and volume of 32 atm and 1.0 L to a final volume of 4.0 L. The initial t

emperature of the gas is 300 K. What are the final pressure and temperature of the gas and how much work is done by the gas if the expansion is (a) isothermal, and (b) isobaric
Physics
1 answer:
Hitman42 [59]3 years ago
8 0

Answer:

(a) P2 = 8 atm, T2 = 300 K, W = 3458.32 J

(b) P2 = 32 atm, T2 = 1200 K, W = 9696 J

Explanation:

P1 = 32 atm

V1 = 1 L

V2 = 4 L

T1 = 300 K

(a) When the process is isothermal

The temperature remains constant, so the final temperature, T2 = 300 k

Use

P1 x V1 = P2 x V2

32 x 1 = P2 x 4

P2 = 8 atm

So, the final pressure is 8 atm and the final temperature is 300 K

Work done in isothermal expansion is given by

W = 2.303 RT log\left ( \frac{V_{2}}{V_{1}} \right )

W = 2.303 \times 8.314\times 300\times  log 4

W = 3458.32 J

(b) When the process is isobaric

the pressure remains constant, so the final pressure, P2 = 32 atm

Use

V1 / T1 = V2/ T2

1 / 300 = 4 / T2

T2 = 1200 K

Work done in isobaric process

W = P (V2 - V1) = 32 x (4 - 1) = 96 atm L

W = 96 x 1.01 x 10^5 x 10^-3 = 9696 J

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If you heat a gas you give the molecules more energy so they move faster. This means more impacts on the walls of the container and an increase in the pressure. Conversely if you cool the molecules down they will slow and the pressure will be decreased.

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Review. An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT . The angular momen
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The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

The angular momentum(L) of an electron moving in a circular path is given by the formula,

L = mvr ........(i)

We know that the radius of the path of an electron in a magnetic field is

r = mv/qB

Putting this value in equation (i),

L = mv x mv/qB

or L = (mv)^2/qB

Putting the given values in the above equation,

4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3

v comes out to be 8.88 x 10^7 m/s.

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5 0
1 year ago
A stone is dropped from a bridge abd it hits the water 2.2 seconds later how high is the bridge above the water
Bess [88]

Answer:

h = 23.716 m

Explanation:

Given that,

The time taken by the stone to hit the water is, t = 2.2 s

Height of the bridge above the ground, h = ?

The distance that the body will fall through the time is given by the formula

                                S = 1/2 gt²  m

Where,        

                              g - acceleration due to gravity

Substituting the values in the above equation

                               S = 1/2 x 9.8 m/s² x (2.2 s)²

                                  = 23.716  m

Therefore, the height of the bridge from the surface of the water is h = 23.716  m

8 0
3 years ago
Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m
maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

T is the tension in the string

\mu is the mass per unit length

For the string in the problem,

L = 30.0 m

\mu=9.00\cdot 10^{-3} kg/m

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz

The next frequencies (harmonics) are given by

f_n = nf

with n being an integer number and f being the fundamental frequency.

So we get:

f_2 = 2 (0.786 Hz)=1.572 Hz

f_3 = 3 (0.786 Hz)=2.358 Hz

f_4 = 4 (0.786 Hz)=3.144 Hz

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