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mezya [45]
3 years ago
11

I need to figure out A but i’m not sure how

Physics
1 answer:
Damm [24]3 years ago
4 0

Answer:

cc

Explanation:

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An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:
Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

\texttt{ }

<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

\large {\boxed {PV = nRT} }

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<h3>Step A:</h3>

<em>Ideal gas is allowed to expand isothermally:</em>

P_1V_1 = P_2V_2

5.00 \times 2.00 = 3.00 \times V_2

V_2 = 10 \div 3

V_2 = 3\frac{1}{3} \texttt{ L}

\texttt{ }

<em>Next we will calculate the work done on the gas:</em>

W_A = -P_2(V_2 - V_1)

W_A = -3.00(3\frac{1}{3} - 2.00)

W_A = \boxed{-4 \texttt{ L.atm}}

\texttt{ }

<h3>Step B:</h3>

<em>Using the same method as above:</em>

P_2V_2 = P_3V_3

3.00 \times 3\frac{1}{3} = 2.00 \times V_3

V_3 = 10 \div 2

V_3 = 5 \texttt{ L}

\texttt{ }

<em>Next we will calculate the work done on the gas:</em>

W_B = -P_3(V_3 - V_2)

W_B = -2.00(5 - 3\frac{1}{3})

W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}

\texttt{ }

<em>Finally we could calculate the total work done and heat added as follows:</em>

W = W_A + W_B

W = -4 + (-3\frac{1}{3})

W = -7\frac{1}{3} \texttt{ L.atm}

W = -7\frac{1}{3} \times 101.33 \texttt{ J}

\boxed{W \approx -743 \textt{ J}}

\texttt{ }

\Delta U = Q + W

0 = Q + (-743)

\boxed{Q = 743 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Minimum Coefficient of Static Friction : brainly.com/question/5884009
  • The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

5 0
4 years ago
Please solve for 15 points. Please don’t input a link.
melomori [17]

Answer:

a). Single replacement.

Explanation:

Because one element replaces another element in a compound

6 0
3 years ago
a car collides with a wall. compare the forces exerted by the car on the wall and the wall on the car​
NeTakaya

Mass multiplied by acceleration produces force.

The acceleration is (v - 0)/t in this situation, where t seems to be the time it takes automobile A to come to a stop. According to Newton's third law of motion, the automobile produces this turning force of the wall, however the wall, which really is static and indestructible, forces an equal force back on the car.

According to Newton's third law, each action has an equal and opposite response. On this basis, you may deduce that a car driving into a wall would exert force on the wall. However, since the wall did not move, the automobile receives an equivalent force, causing it to collapse.

<h3>Learn more:</h3>

brainly.com/question/13952508?referrer=searchResults

5 0
2 years ago
Read 2 more answers
What is the velocity of a car that travels -50 meters in 5 seconds then 10 meters in 5 seconds
diamong [38]

Answer:

2 m/s

Explanation:

The first part of the question the car is going in reverse or negative along the x axis. Then the second part the car is moving forward along the x axis. So the car would only have velocity in the current direction of movement. So our equation for velocity is as follows.

v = d/t

v = 10 m/5 s

v = 2 m/s

3 0
3 years ago
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re
spayn [35]

The magnitude of the electric field can be calculated using the equation

E = \frac{F}{q}, where F is the Force acting on the proton and q is the charge of the proton.

We know that the charge of the proton is q = 1.6 x 10^{-19}  C

We have to calculate the Force first.

We know that F = ma from Newton's 2nd law, where m is the mass of the proton and a is the acceleration.

We know that the mass of the proton m = (1.67) X 10^{-27}  kg

So it turns out that we have to calculate acceleration before anything else.

In order to calculate the acceleration, we make use of the following data from the question:

Initial Velocity of the proton V_{i}  = 0

Distance traveled  D = 1.70 cm = 0.017 m

Time taken for the travel between the plates t = (1.48) X 10^{-6}  s

Acceleration a = ?

Using the equation, D = V_{i}t + \frac{1}{2} at^{2}, we get

Knowing that initial velocity is 0, the equation reduces to D = \frac{1}{2}at^{2}

Rearranging the equation so as to make a the subject of the formula, we have

a = \frac{2D}{t^{2} }

Plugging in the numbers and simplifying gives us a = 1.5 x 10^{10}   m/s^{2}

We can now calculate the Force using F = ma

Plugging in the known values, we get F = 2.5 x 10^{-17}  N

Using this, we can calculate E through the equation E = \frac{F}{q}

Plugging the numbers and simplifying gets us E = 156.25 N/C

Thus, the magnitude of the electric field between the plates of the capacitor is 156.25 N/C


B) To calculate the Final Velocity of the proton, we can make use of the equation

V_{f}  = V_{i}  + at

Plugging the numbers in and simplifying gets us V_{f}  = (2.22)  *  10^{4}  m/s

5 0
4 years ago
Read 2 more answers
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