Question

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:
a.against a constant external pressure of 3.00 atm, followed by
b.against a constant external pressure of 2.00 atm. calculate q and w. (101.33 j = 1 l atm)

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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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

Answer 2

The value of Q(net heat transfer) and W(work done) are;

Q = 0

Q = 0W = -743 J

We are given;

Initial volume; V1 = 2 L

Initial pressure of the gas; P1 = 5 atm

A) Since the gas expands isothermally, then we5 can use Boyles law to get;

P1V1 = P2V2

Here P2 = P_ext = 3 atm

Thus;

V2 = P1 × V1/P2

V2 = 5 × 2/3

V2 = 3.33 L

Formula for with external pressure of 3 atm is;

W = -P_ext(V2 - V1)

W1 = -3(3.33 - 2)

W1 = -4 atm.L

B) Similarly;

P2V2 = P3V3

P3 = P_ext,3 = 2 atm

V3 = P2 × V2/P3

V3 = (3 × 3.33)/2

V3 = 5 L

Thus;

W2 = -2(5 - 3.33)

W2 = -3.33 atm.L

Total workdone is;

W = W1 + W2

W = -4 - 3.33

W = -7.33 atm.L

We are told that;

1 atm.L = 101.33 J

Thus; W = -7.33 × 101.33

W ≈ -743 J

Now, from first law of thermodynamics, we know that;

ΔU = Q - W

In this, Q the net heat transfer is 0 because the gas expands isothermally.

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