Question

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

Answer 1

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

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Further explanation

The Ideal Gas Law that needs to be recalled 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)

Let us now tackle 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 added to the gas = Q = ?

Solution:

Step A:

Ideal gas is allowed to expand 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 calculate the work done 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:

Using the same method as above:

$P_2V_2 = P_3V_3$

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

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

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Next we will calculate the work done 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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Finally we could calculate the total work done and heat added 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 : 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

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure