Initial and Final States:
- Initial pressure (P1) = 5 atm
- Final pressure (P2) = 1 atm
- Initial volume (V1) = ?
- Final volume (V2) = ?
- Initial temperature (T1) = 25°C = 298 K
- Final temperature (T2) = ?

Calculating the Final Volume:
We can use Boyle's Law to find the final volume (V2) of the nitrogen gas.
According to Boyle's Law: P1 * V1 = P2 * V2
Substituting the values, we get:
5 atm * V1 = 1 atm * V2
V2 = (5 atm / 1 atm) * V1
V2 = 5 * V1

Calculating the Final Temperature:
To calculate the final temperature (T2) of the nitrogen gas, we can use the ideal gas law equation:
P1 * V1 / T1 = P2 * V2 / T2
Substituting the known values, we get:
5 atm * V1 / 298 K = 1 atm * 5 * V1 / T2
Simplifying the equation, we find:
T2 = (5 atm * 298 K) / (5 atm)
T2 = 298 K

Entropy Change Calculation:
The entropy change (ΔS) in an ideal gas can be calculated using the equation:
ΔS = nR ln(V2/V1) + nCv ln(T2/T1)
where:
- n is the number of moles of the gas
- R is the ideal gas constant (8.314 J/(mol·K))
- ln represents the natural logarithm
- Cv is the molar heat capacity at constant volume

Calculating the Number of Moles:
To find the number of moles (n) of nitrogen gas, we can use the equation:
n = mass / molar mass
Given that the mass of nitrogen is 200 g and the molar mass of nitrogen is approximately 28 g/mol, we can calculate:
n = 200 g / 28 g/mol
n ≈ 7.14 mol

Calculating the Entropy Change:
Substituting the known values into the entropy change equation:
ΔS = (7.14 mol * 8.314 J/(mol·K)) ln((5 * V1)/V1) + (7.14 mol * Cv) ln(298 K/25°C)
Since nitrogen is an ideal gas, its molar heat capacity at constant volume (Cv) is approximately 20.8 J/(mol·K).
ΔS = 59.37 J/K + (7.14 mol * 20.8 J/(mol·K)) ln(298 K/25°C)
Simplifying and converting the temperature from Celsius to Kelvin:
ΔS ≈ 59.37 J/K + 148.51 J/K
ΔS ≈ 207.88 J/K

Conclusion:
The entropy change in the process of compressing 200 g of nitrogen gas from 5 atm to 1 atm at 25°C is approximately 207.88 J/K. This calculation considers the behavior