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Julli [10]
3 years ago
15

A certain gas is present in a 15.0 l cylinder at 2.0 atm pressure. if the pressure is increased to 4.0 atm the volume of the gas

decreases to 7.5 l . find the two constants ki, the initial value of k, and kf, the final value of k, to verify whether the gas obeys boyle's law by entering the numerical values for ki and kf in the space provided.
Physics
1 answer:
raketka [301]3 years ago
4 0

In the first condition: P1 = 2.0 atm V1 = 15.0 L P1V1 = Ki 2 * 15 = Ki Ki = 30

In the second condition Pt = 4.0 atm Vt = 7.5L Pt * Vt = Kt 4.0 * 7.5 = Kt Kt = 30

Conclusion: Ki = Kt. Since the two values of k are the same. Therefore, the gas obeys Boyle's Law. Take note Boyle’s law is an ideal gas law which tells us that the volume of an ideal gas is contrariwise proportional to its absolute or total pressure at a constant temperature.
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A pendulum makes 60 vibrations in 15 secs what’s its frequency
o-na [289]

Answer:

4 hertz

Explanation:

The defination of freqyency = the total no of cycle made by a wave in one second .

so,

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A train travels 77 kilometers in 4 hours, and then 76 kilometers in 2 hours. What is its average speed?
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4 0
3 years ago
The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0
3 years ago
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