One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at constant volume, Cv=1000 kj/kg.K at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.:
Question
One kg of an ideal gas (gas constant, R = 400 J/kg.K; specific heat at constant volume, Cv=1000 kj/kg.K at 1 bar, and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100kJ of work is done on the system by a stirrer. The increase in entropy of the system is _________ J/K.:
Solution
To solve this problem, we need to use the first and second laws of thermodynamics.
Step 1: Calculate the initial internal energy of the system. The internal energy (U) of an ideal gas is given by the equation U = Cv*T, where Cv is the specific heat at constant volume and T is the temperature.
U1 = Cv*T1 = 1000 kJ/kg.K * 300 K = 300,000 kJ/kg = 300,000,000 J/kg
Step 2: Calculate the final internal energy of the system. The work done on the system increases its internal energy. The work done on the system is given as 100 kJ = 100,000 J.
U2 = U1 + W = 300,000,000 J/kg + 100,000 J = 300,100,000 J/kg
Step 3: Calculate the final temperature of the system. We can find the final temperature by rearranging the equation for internal energy to solve for T.
T2 = U2 / Cv = 300,100,000 J/kg / 1000 kJ/kg.K = 300,100 K
Step 4: Calculate the change in entropy. The change in entropy (ΔS) for an adiabatic process in an ideal gas is given by the equation ΔS = Cv*ln(T2/T1), where ln is the natural logarithm.
ΔS = Cv*ln(T2/T1) = 1000 kJ/kg.K * ln(300,100/300) = 1000 * 0.000333 = 0.333 kJ/K = 333 J/K
So, the increase in entropy of the system is 333 J/K.
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