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zlopas [31]
3 years ago
9

A gas is inside a cylinder fitted with a piston. The gas and its surroundings are both at 20 ∘C and the gas is compressed as the

piston moves and decreases the cylinder volume. The compression takes place slowly enough to allow the gas temperature to stay at 20 ∘C and the work done on the gas by the compression is 1.9 × 103 J. What is the change in entropy of the gas
Physics
1 answer:
Alex73 [517]3 years ago
8 0

Answer:

-  6.48 J K⁻¹

Explanation:

Temperature of gas = 20°C

= 273 +20 = 293 K.

As temperature is fixed , there will be no change in gas internal energy

Δ E = 0

From the relation

Δ Q =  Δ E+ Δ W

Work done on the gas is 1.9 x 10³

Δ W = - 1.9 x 10³ J

Δ Q = - 1.9 x 10³ J

Δ S =  Δ Q / T

=\frac{- 1.9 \times 10^3 }{293}

= -  6.48 J K⁻¹

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Answer:

This will require 266.9 of heat energy.

Explanation:

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c

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Here is a source of values of

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Once you have all that, this is the equation:

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Assume that you can drive at a constant speed of 100 kilometers per hour. suppose you started driving from the sun. how long wou
belka [17]
The Sun is 149.6 million kilometers from the earth.
There are 8760 hours in a year. 
876000 km are traveled in a year
It would take 170.776 years to reach the sun, or 171 years rather
4 0
3 years ago
Problem 4: a long wire carries current towards east. a positive charge moves westward and just north from the wire. what is the
Alex787 [66]
The direction of the force experienced by the positive charge is upward.

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3 years ago
calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s
IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m  * 1609.344 m

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next we will express the distance between the earth and the sun

r = \frac{a(1-E^2)}{1+Ecos\beta }   --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

\beta = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

v^2 = \frac{4\pi^2 }{r_{c} }   ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

(V^2 = (\frac{4\pi^{2}  }{149.626*10^9})

therefore : V = 1.624* 10^-5 m/s

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3 years ago
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