Question

A gas is compressed from an initial volume of 5.75 L to a final volume of 1.23 L by an external pressure of 1.00 atm. During the compression the gas releases 128 J of heat. What is the change in internal energy of the gas

Answer 1

Answer:

The internal energy change is 330.01 J

Explanation:

Given

$v_{1}$ the initial volume = 5.75 L

$v_{2}$ the final volume = 1.23 L

$P_{e}$ is the external pressure = 1.00 atm

q the heat energy removed = -128 J (since is removed from the system)

expansion against a constant external pressure is an example of an irreversible pathway, here pressure in is greater than pressure out and can be obtained thus;

W = -$P_{e}$ΔV

W = -1.00 x(1.23 - 5.75)

W =  -1.00 x -4.52

W = 4.52 L atm

converting to joules we have

W = 4.52 L atm x 101.33 J/ L atm = 458.01 J

The internal energy change during compression can be calculated thus;

ΔU = q + W

ΔU = -128 J + 458.01 J

ΔU = 330.01 J

Therefore the internal energy change is 330.01 J

Answer 2

The change in internal energy of the gas is equal to 586.01 Joules.

Given the following data:

• Initial volume = 5.75 Liters.

• Final volume of 1.23 Liters.

• Pressure = 1 atm

• Quantity of heat energy = 128 Joules.

To determine the change in internal energy of the gas:

First of all, we would calculate the work energy of this gas.

Mathematically, the work energy by a system is given by the formula:

$Work = -P \delta V$

Where:

• P is the pressure.

• $\delta V$ is the change in volume.

Substituting the given parameters into the formula, we have;

$Work\;energy = -1(1.23-5.75)\\\\Work\;energy = -1(-4.52)$

Work energy = 4.52 Latm.

Conversion:

1 Latm = 101.33 Joules

4.52 Latm = $4.52 \times 101.33 = 458.01 \;Joules$

Now, we can determine the change in internal energy by using this formula:

$\Delta E = q + w\\\\\Delta E = 128 + 458.01\\\\\Delta E =586.01 \;Joules$

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