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Nitrogen gas (molecular weight 28) is enclosed in a cylinder by a piston, at the initial condition of 2 bar, 298 K and 1 m3. In a particular process, the gas slowly expands under isothermal condition, until the volume becomes 2m3. Heat exchange occurs with the atmosphere at 298 K during this process.
The entropy change for the Universe during the process in kJ/K is
- a)0.4652
- b)0.0067
- c)0
- d)–0.6711
Correct answer is option 'C'. Can you explain this answer?
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Entropy Change for the Universe during the Process
Given data:
Initial conditions:
- Pressure (P1) = 2 bar
- Temperature (T1) = 298 K
- Volume (V1) = 1 m^3
Final conditions:
- Volume (V2) = 2 m^3
- Temperature (T2) = 298 K
To calculate the entropy change for the universe during the process, we need to consider both the system (nitrogen gas) and the surroundings (atmosphere).
1. Entropy Change of the System (Nitrogen Gas)
The entropy change of the system can be calculated using the ideal gas equation and the definition of entropy change.
The ideal gas equation is given by:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
Since the process is isothermal, the temperature remains constant. Therefore, the ideal gas equation can be written as:
P1V1 = P2V2
To determine the number of moles, we can use the molecular weight of nitrogen gas:
n = m/M
Where:
m = mass of nitrogen gas
M = molecular weight of nitrogen gas
The entropy change of the system is given by:
ΔS_system = nR ln(V2/V1)
2. Entropy Change of the Surroundings (Atmosphere)
The entropy change of the surroundings can be calculated using the heat transfer equation and the definition of entropy change.
The heat transfer equation is given by:
Q = mCΔT
Where:
Q = heat transfer
m = mass of the surroundings
C = specific heat capacity of the surroundings
ΔT = change in temperature
Since the process is isothermal, the temperature remains constant. Therefore, the heat transfer equation simplifies to:
Q = 0
The entropy change of the surroundings is given by:
ΔS_surroundings = -Q/T
3. Entropy Change for the Universe
The entropy change for the universe is the sum of the entropy changes of the system and the surroundings:
ΔS_universe = ΔS_system + ΔS_surroundings
Given that the heat transfer between the system and the surroundings is zero, the entropy change for the universe simplifies to:
ΔS_universe = ΔS_system
To calculate the entropy change, we can substitute the values into the equation:
ΔS_universe = nR ln(V2/V1)
Substituting the values:
n = m/M = (mass of nitrogen gas)/(molecular weight of nitrogen gas) = 1/28
R = gas constant = 8.314 J/(mol·K)
V2/V1 = 2/1 = 2
ΔS_universe = (1/28) * (8.314 J/(mol·K)) * ln(2)
ΔS_universe = 0.297 J/K
Converting to kJ/K:
ΔS_universe = 0.297 J/K * (1 kJ/1000 J)
ΔS_universe ≈ 0.0003 kJ/K
Therefore, the correct answer is option 'C' - 0.
Given data:
Initial conditions:
- Pressure (P1) = 2 bar
- Temperature (T1) = 298 K
- Volume (V1) = 1 m^3
Final conditions:
- Volume (V2) = 2 m^3
- Temperature (T2) = 298 K
To calculate the entropy change for the universe during the process, we need to consider both the system (nitrogen gas) and the surroundings (atmosphere).
1. Entropy Change of the System (Nitrogen Gas)
The entropy change of the system can be calculated using the ideal gas equation and the definition of entropy change.
The ideal gas equation is given by:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
Since the process is isothermal, the temperature remains constant. Therefore, the ideal gas equation can be written as:
P1V1 = P2V2
To determine the number of moles, we can use the molecular weight of nitrogen gas:
n = m/M
Where:
m = mass of nitrogen gas
M = molecular weight of nitrogen gas
The entropy change of the system is given by:
ΔS_system = nR ln(V2/V1)
2. Entropy Change of the Surroundings (Atmosphere)
The entropy change of the surroundings can be calculated using the heat transfer equation and the definition of entropy change.
The heat transfer equation is given by:
Q = mCΔT
Where:
Q = heat transfer
m = mass of the surroundings
C = specific heat capacity of the surroundings
ΔT = change in temperature
Since the process is isothermal, the temperature remains constant. Therefore, the heat transfer equation simplifies to:
Q = 0
The entropy change of the surroundings is given by:
ΔS_surroundings = -Q/T
3. Entropy Change for the Universe
The entropy change for the universe is the sum of the entropy changes of the system and the surroundings:
ΔS_universe = ΔS_system + ΔS_surroundings
Given that the heat transfer between the system and the surroundings is zero, the entropy change for the universe simplifies to:
ΔS_universe = ΔS_system
To calculate the entropy change, we can substitute the values into the equation:
ΔS_universe = nR ln(V2/V1)
Substituting the values:
n = m/M = (mass of nitrogen gas)/(molecular weight of nitrogen gas) = 1/28
R = gas constant = 8.314 J/(mol·K)
V2/V1 = 2/1 = 2
ΔS_universe = (1/28) * (8.314 J/(mol·K)) * ln(2)
ΔS_universe = 0.297 J/K
Converting to kJ/K:
ΔS_universe = 0.297 J/K * (1 kJ/1000 J)
ΔS_universe ≈ 0.0003 kJ/K
Therefore, the correct answer is option 'C' - 0.
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Nitrogen gas (molecular weight 28) is enclosed in a cylinder by a piston, at the initial condition of 2 bar, 298 K and 1 m3. In a particular process, the gas slowly expands under isothermal condition, until the volume becomes 2m3. Heat exchange occurs with the atmosphere at 298 K during this process.The entropy change for the Universe during the process in kJ/K isa)0.4652b)0.0067c)0d)–0.6711Correct answer is option 'C'. Can you explain this answer?
Question Description
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