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harina [27]
3 years ago
5

Problem 11.010 IS A rigid tank contains 1 kg of oxygen (O2) at p1 = 30 bar, T1 = 180 K. The gas is cooled until the temperature

drops to 150 K. Determine the volume of the tank, in m3, and the final pressure, in bar, using the:
Physics
1 answer:
aliina [53]3 years ago
7 0

Answer:

Explanation:

The gas law relation is

PV = nRT where n is no of moles of gas

given 1 kg of oxygen

n = 1000 / 32 = 31.25 moles

P = 30 bar = 30 x 10⁵ Pa

V =?

T = 180 K

R = 8.32

Substituting the values

30 x 10⁵ x V = 31.25 x 8.32 x 180

V = 1560 x 10⁻⁵ m³

= 1.56 x 10⁻² m³

When volume remains constant , pressure is proportional to temperature

P₂ / P₁ = T₂ / T₁

P₂ =  ( 150 / 180 ) x 30

= 25 Bar.

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describe the trends of the elements including boiling/melting points and conductivity in relation to the periodic table
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modern quantum mechanics explains these periodic trends in properties in terms of electron shells. the filling of each shell corresponds to a row in the table

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Lin suffers from pain in her right wrist. Several doctors check her wrist regularly over a period of time. Doctors' notes and
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3 years ago
A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates
Novosadov [1.4K]

Answer:

4334.4 J

Explanation:

Work done equals to kinetic energy change

KE=½mv²

Change in KE is given by

∆KE=½m(v²-u²)

Where m is mass of water-skier, KE is kinetic energy, ∆KE is the change in kinetic energy, v is final velocity and u is initial velocity.

Substituting 72 kg for m, 12.1 m/s for v and 5.10 m/s for u then

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3 years ago
In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e
Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_1-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_3)}{Q_1}}

also,

\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}

or

\frac{T_3}{T_1}=\frac{Q_3}{Q_1}

thus,

\eta=1-\frac{(T_3)}{T_1}}

thus,

\eta=1-\frac{(240\ K)}{500\ K}}

or

\eta=0.52

or

Efficiency = 52%

8 0
4 years ago
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