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marishachu [46]
3 years ago
12

A gas is compressed at a constant pressure of 0.800 atm from 12.00 L to 3.00 L. In the process, 390 J of energy leaves the gas b

y heat.
a) What is the work done on the gas?
b) What is the change in its internal energy?
Physics
1 answer:
Andru [333]3 years ago
5 0

Answer:

a)W= - 720 J

b)ΔU= 330 J

Explanation:

Given that

P = 0.8 atm

We know that 1 atm = 100 KPa

P = 80 KPa

V₁ = 12 L = 0.012 m³       ( 1000 L = 1 m³)

V₂ = 3 L = 0.003 m³

Q= - 390 J ( heat is leaving from the system )

We know that work done by gas given as

W = P (V₂ -V₁ )

W= 80 x ( 0.003 - 0.012 ) KJ

W= - 0.72 KJ

W= - 720 J    ( Negative sign indicates work done on the gas)

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in the internal energy

Now by putting the values

- 390 = - 720 + ΔU

ΔU= 720 - 390  J

ΔU= 330 J

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The maximum amount of work performed is

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Explanation:

The efficiency of a real heat engine is given by the equation:

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where

T_C is the temperature of the cold reservoir

T_H is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

\eta = \frac{W_{max}}{Q_H}

where

W_{max} is the maximum work done

Q_H is the heat absorbed from the hot reservoir

Q_H can be written as

Q_H=W_{max}+Q_C

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By combining eq.(1) and (2) we get

1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}

And re-arranging the equation and solving for W_{max}, we find

W_{max}=\frac{T_H-T_C}{T_C}Q_C

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3 years ago
Why is a minimum of three seismic stations needed to find the epicenter of an earthquake?
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A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of
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Given :

A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg.

To Find :

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Since no external force is acting in the system, therefore total energy will be conserved.

Initial kinetic energy of the object = Energy stored in spring

K.E _i = \dfrac{kx^2}{2}\\\\K.E_i = \dfrac{1730\times 0.136^2}{2}\\\\K.E_i = 16\ J

Also, initial potential energy is 0.

Now,

K.E_i + P.E_i = K.E_f + P.E_f\\\\16 + 0 = \dfrac{1.72\times 2.45^2}{2}+ mgh\\\\mgh =16 - 5.16\\\\h = \dfrac{16 - 5.16}{1.72 \times 9.8}\\\\h = 0.64\ m

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3 years ago
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Answer:

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Given that a small beaker has 50 mL of water in it. A small frog with a mass of 20 grams is dropped into the beaker. The water level rises to 79mL.

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79 - 50 = 29mL

Convert it to litre by dividing it by 1000

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Since 1L = 1000 cm^3

Convert it to cm^3 by multiplying it by 1000

0.029 × 1000 = 29 cm^3

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Substitute the mass and volume into the formula

Density = 20/29

Density = 0.6896 g/cm3

Therefore, the density of the frog is 0.69 g/cm^3 approximately

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Answer:

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