Calculation of Entropy Change

Given:
Initial volume (Vi) = 1 L
Final volume (Vf) = 10 L
Temperature (T) = 27 °C = 300 K
Mass of N2 gas (m) = 2.8 g

Step 1: Calculate the moles of N2 gas
The number of moles (n) can be calculated using the formula:
n = mass / molar mass

The molar mass of N2 gas is:
Molar mass (M) = 28 g/mol

Substituting the values:
n = 2.8 g / 28 g/mol
n = 0.1 mol

Step 2: Calculate the change in volume
The change in volume (ΔV) can be calculated by subtracting the initial volume from the final volume:
ΔV = Vf - Vi
ΔV = 10 L - 1 L
ΔV = 9 L

Step 3: Calculate the gas constant (R)
The gas constant (R) is given by:
R = 8.314 J/(mol·K)

Step 4: Calculate the entropy change (ΔS)
The entropy change (ΔS) can be calculated using the formula:
ΔS = nRln(Vf/Vi)

Substituting the values:
ΔS = 0.1 mol * 8.314 J/(mol·K) * ln(10 L / 1 L)
ΔS = 0.1 * 8.314 * ln(10)

Using ln(10) ≈ 2.3026:
ΔS ≈ 0.1 * 8.314 * 2.3026
ΔS ≈ 1.917 J/K

Explanation:
The entropy change of a system is a measure of the disorder or randomness of the system. In this case, we have an isothermal (constant temperature) and reversible expansion of N2 gas.

In the first step, we calculate the number of moles of N2 gas using the given mass and molar mass. We find that there are 0.1 moles of N2 gas.

In the second step, we calculate the change in volume by subtracting the initial volume from the final volume. The change in volume is 9 L.

In the third step, we calculate the gas constant, which is a constant value used in thermodynamic calculations. The gas constant is 8.314 J/(mol·K).

Finally, in the fourth step, we calculate the entropy change using the formula ΔS = nRln(Vf/Vi). Substituting the known values, we find that the entropy change is approximately 1.917 J/K.

Therefore, the entropy change when 2.8g of N2 gas expands isothermally and reversibly from an initial volume of 1L to a final volume of 10L at 27 °C is approximately 1.917 J/K.

1.91 J/K ?