The Stern-Gerlach experiment of 1922 established that each individual

A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a friction

marissa [1.9K]

Answer:

Part a: The total amount of energy transfer by the work done is 54.81 kJ.

Part b: The total amount of energy transfer by the heat is 54.81 kJ

Explanation:

Mass of Carbon Dioxide is given as m1=3 kg

Pressure is given as P1=3 bar =300 kPA

Volume is given as V1=0.5 m^3

Pressure in tank 2 is given as P2=2 bar=200 kPa

T=290 K

Now the Molecular weight of $CO_2$ is given as

M=44 kg/kmol

the gas constant is given as

$R=\frac{\bar{R}}{M}\\R=\frac{8.314}{44}\\R=0.189 kJ/kg.K$

Volume of the tank is given as

$V=\frac{mRT}{P_1}\\V=\frac{3 \times 0.189 \times 290}{300 }\\V=0.5481 m^3$

Final mass is given as

$m_2=\frac{P_2V}{RT}\\m_2=\frac{200\times 0.5481}{0.189\times 290}\\m_2=2 kg$

Mass of the CO2 moved to the cylinder

$m=m_1-m_3\\m=3-2=1 kg$

The initial mass in the cylinder is given as

$m_{(cyl)_1}=\frac{P_{(cyl)_1}V_1}{RT}\\m_{(cyl)_1}=\frac{200\times 0.5}{0.189 \times 290}\\m_{(cyl)_1}=1.82 kg$

The mass after the process is

$m_{(cyl)_2}=m_{(cyl)_1}+m\\m_{(cyl)_2}=1.82+1\\m_{(cyl)_2}=2.82\\$

Now the volume 2 of the cylinder is given as

$V_{(cyl)_2}=\frac{m_{(cyl)_2}RT}{P_2}\\m_{(cyl)_2}=\frac{2.82\times 0.189\times 290}{200}\\m_{(cyl)_1}=0.774 m^3$

Part a:

So the Work done is given as

$W=P(V_2-V_1)\\W=200(0.774-0.5)\\W=54.81 kJ$

The total amount of energy transfer by the work done is 54.81 kJ.

Part b:

The total energy transfer by heat is given as

$Q=\Delta U+W\\Q=0+W\\Q=54.81 kJ$

As the temperature is constant thus change in internal energy is 0.

The total amount of energy transfer by the heat is 54.81 kJ

7 0

3 years ago

A 1.00- and a 2.00- resistor are in parallel. What is the equivalent single resistance?

pochemuha

Answer: R=2/3 ohms

Explanation:

4 0

3 years ago

A water balloon was dropped from a high window and struck its target 1.1 seconds later. If the balloon left the person's hand at

valkas [14]

-5.5 is the answer to the question

8 0

3 years ago

Read 2 more answers

The half life of radium 222 is 38 seconds if you had a 12 gram sample how much would be left after 72 seconds

poizon [28]

3.2 grams The first thing to do is calculate how many half lives have expired. So take the time of 72 seconds and divide by the length of a half life which is 38 seconds. So 72 / 38 = 1.894736842 So we're over 1 half life, but not quite 2 half lives. So you'll have something less than 12/2 = 6 grams, but more than 12/4 = 3 grams. The exact answer is done by dividing 12 by 2 raised to the power of 1.8947. So let's calculate 2^1.8947 power = 12 g / (e ^ ln(2)*1.8947) = 12 g / (e ^ 0.693147181 * 1.8947) = 12 g / (e ^ 1.313305964) = 12 g / 3.718446464 = 3.227154167 g So rounded to 2 significant figures gives 3.2 grams.

5 0

4 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

morpeh [17]

Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron, $u_2=0$