For a process where T1 = 300K, P1 = 100 kPa, and where T2 = 500K, P2 = 650 kPa, the change in entropy is mathematically given as
S1-S2=-0.0288KJ/Kgk
Generally, the equation for the entropy is mathematically given as
S1-S2=mcp*lnt2-t1-Rlnt2/t1
Therefore
S1-S2=1+1.004*ln(500/300)-0.287*ln(650/100)
S1-S2=-0.0288KJ/Kgk
In conclusion, Entropy
S1-S2=-0.0288KJ/Kgk
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Complete Question
Air in a piston-cylinder assembly undergoes a process from state 1, where T1 = 300K, P1 = 100 kPa, to state 2, where T2 = 500K, P2 = 650 kPa. Using the ideal gas model for air, determine the change in entropy between the two states, in kJ/kg-K
The change in entropy between the two states, in a piston cylinder assembly, undergoes a process from state 1 where t1=300k is 2.36 kJ/kg K.
The change in entropy can be find out using the following formula,
[tex]S1-S2=s^oT_1-R\ln \dfrac{p_2}{p_1}[/tex]
Here, (p1 and p2) are the pressure at state 1 and 2 and R is the gas constant. The value of R is,
R=(8.314/28.97).
Air in a piston-cylinder assembly undergoes a process from state 1, where T1 = 300 K, p1 = 100 kPa, to state 2. where T2 = 500 K, p2 = 650 kPa.
Put these values,
[tex]S_2-S_1=1.70203_1-\dfrac{8.314}{28.97}\ln \dfrac{10\text { bar}}{1\text { bar}}\\S_2-S_1=2.36\rm\; kJ/kg K[/tex]
Thus, the change in entropy between the two states, in a piston cylinder assembly undergoes a process from state 1 where t1=300k is 2.36 kJ/kg K.
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