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katen-ka-za [31]
3 years ago
9

10 kg of R-134a fill a 1.115-m^3 rigid container at an initial temperature of -30∘C. The container is then heated until the pres

sure is 200 kPa. Determine the final temperature an the initial pressure. Answers: 14.2∘C, 84.43 kPa
Physics
1 answer:
Sidana [21]3 years ago
4 0

Answer : The final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa

Explanation :

First we have to calculate the initial pressure of gas.

As we know that, R-134a is 1,1,1,2-Tetrafluoroethane.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT

where,

P = pressure of gas = ?

V = volume of gas = 1.115m^3

T = temperature of gas = -30^oC=273+(-30)=243K

R = gas constant = 8.314m^3Pa/mole.K

w = mass of gas = 10 kg = 10000 g

M = molar mass of R-134a gas = 102.03 g/mole

Now put all the given values in the ideal gas equation, we get:

P\times 1.115m^3=\frac{10000g}{102.03g/mole}\times (8.314m^3Pa/mole.K)\times (243K)

P=177587.9687Pa=177.6kPa

Thus, the initial pressure of gas is, 177.6 kPa

Now we have to calculate the final temperature of gas.

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

or,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1 = initial pressure of gas = 177.6 kPa

P_2 = final pressure of gas = 200 kPa

T_1 = initial temperature of gas = -30^oC=273+(-30)=243K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{177.6kPa}{243K}=\frac{200kPa}{T_2}

T_2=273.6K

Thus, the final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa

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⇒  \frac{v}{u} =\frac{h_{1}}{h_{0}}

⇒  \frac{1.94}{25}=\frac{{h_{1}}}{15}

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Take the mechanical equivalent of heat as 4 J/cal. A 10-g bullet moving at 2000 m/s plunges
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Assuming that all the bullet’s energy heats the paraffin, its final temperature is 27.1 degree C. The correct option is D.

<h3>What is temperature?</h3>

Temperature is the degree of hotness and coldness of the material.

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E = 1/2 x 10 x 10⁻³ x (2000)²

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A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of
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Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height \Delta h depends on the square of the time:

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t,

where

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  • g is the acceleration due to gravity.
  • t is the time for which the ball is in the air.
  • v_0 is the initial vertical velocity of the ball.
  • The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. \Delta h = -0.950\;\text{m}.
  • Gravity pulls objects toward the earth, so g is also negative. g \approx -9.81\;\text{m}\cdot\text{s}^{-2} near the surface of the earth.
  • Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, v_0 = 0.

Solve for t.

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t;

\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2};

\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)};

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What's the initial horizontal velocity of the ball?

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Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}.

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

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3 years ago
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