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3 months ago

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

Physics

1 answer:

Sav [3.1K]3 months ago

5 0

Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

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Additional explanation

The Ideal Gas Law that should be remembered is:

$\large {\boxed {PV = nRT} }$

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

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Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

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

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Next, we will determine the work performed on the gas:

$W_A = -P_2(V_2 - V_1)$

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

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

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Step B:

By utilizing the methodology mentioned earlier:

$P_2V_2 = P_3V_3$

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

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

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Next, we will ascertain the work completed on the gas:

$W_B = -P_3(V_3 - V_2)$

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

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

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Ultimately, we can calculate the total work done and heat supplied as follows:

$W = W_A + W_B$

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

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

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

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

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$\Delta U = Q + W$

$0 = Q + (-743)$

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

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Learn more

Minimum Coefficient of Static Friction:
The Pressure In A Sealed Plastic Container:
Effect of Earth’s Gravity on Objects:

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

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A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in

Maru [3345]

Answer:

The mass of the child is 14.133 kg

Explanation:

According to the principle of linear momentum conservation, we get;

(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂) × v₃ - m₃ × v₄

The negative sign signifies that their velocities were directed oppositely

Considering both the child and the ball are initially at rest, we have;

v₁ = 0 m/s and v₂= 0 m/s

Thus;

0 = m₁ × v₃ - m₂ × v₄

(m₁ + m₂) × v₃ = m₃ × v₄

Where:

m₁ = Mass of the child

m₂ = Mass of the scooter = 2.4 kg

v₃ = Final velocity of the child and scooter = 0.45 m/s

m₃ = Mass of the ball = 2.4 kg

v₄ = Final velocity of the ball = 3.1 m/s

Substituting the known values gives us;

(m₁ + 2.4) × 0.45 = 2.4 × 3.1

(m₁ + 2.4) = 16.533

This implies m₁ + 2.4 = 16.533

Thus, m₁ = 16.533 - 2.4 = 14.133 kg

The mass of the child equals 14.133 kg.

3 0

2 months ago