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Two moles of an ideal gas is expanded isothermally and reversibly from 5 to 1 bar at 298 K. The change in the entropy (in J K–1) of the system is _______
Correct answer is between '26.5,27.0'. Can you explain this answer?
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Two moles of an ideal gas is expanded isothermally and reversibly from...
Solution:
Given, n = 2 moles
T = 298 K
Initial pressure, P1 = 5 bar
Final pressure, P2 = 1 bar
As the process is isothermal, the temperature remains constant throughout the process. Therefore, ΔT = 0.
The change in entropy of the system can be calculated using the following formula:
ΔS = nR ln (V2/V1)
Where,
R is the gas constant (8.314 J K-1 mol-1)
V1 is the initial volume
V2 is the final volume
To calculate V1 and V2, we can use the ideal gas law:
PV = nRT
Where,
P is the pressure
V is the volume
Rearranging the equation, we get:
V = nRT/P
Using this equation, we can calculate the initial and final volumes:
V1 = (2 mol x 8.314 J K-1 mol-1 x 298 K) / (5 bar x 105 Pa/bar) = 9.94 L
V2 = (2 mol x 8.314 J K-1 mol-1 x 298 K) / (1 bar x 105 Pa/bar) = 39.75 L
Substituting the values in the equation for ΔS, we get:
ΔS = (2 mol x 8.314 J K-1 mol-1) x ln (39.75 L / 9.94 L) = 26.8 J K-1
Therefore, the change in entropy of the system is between 26.5 and 27.0 J K-1, which matches the given answer range.
Given, n = 2 moles
T = 298 K
Initial pressure, P1 = 5 bar
Final pressure, P2 = 1 bar
As the process is isothermal, the temperature remains constant throughout the process. Therefore, ΔT = 0.
The change in entropy of the system can be calculated using the following formula:
ΔS = nR ln (V2/V1)
Where,
R is the gas constant (8.314 J K-1 mol-1)
V1 is the initial volume
V2 is the final volume
To calculate V1 and V2, we can use the ideal gas law:
PV = nRT
Where,
P is the pressure
V is the volume
Rearranging the equation, we get:
V = nRT/P
Using this equation, we can calculate the initial and final volumes:
V1 = (2 mol x 8.314 J K-1 mol-1 x 298 K) / (5 bar x 105 Pa/bar) = 9.94 L
V2 = (2 mol x 8.314 J K-1 mol-1 x 298 K) / (1 bar x 105 Pa/bar) = 39.75 L
Substituting the values in the equation for ΔS, we get:
ΔS = (2 mol x 8.314 J K-1 mol-1) x ln (39.75 L / 9.94 L) = 26.8 J K-1
Therefore, the change in entropy of the system is between 26.5 and 27.0 J K-1, which matches the given answer range.
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