Question

Air in a piston cylinder assembly undergoes a process from state 1 where t1=300k

Answer 1

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

What is the change in entropy between the two states, in kJ/kg-K?

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

Answer 2

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.

How to find change in entropy?

The change in entropy can be find out using the following formula,

$S1-S2=s^oT_1-R\ln \dfrac{p_2}{p_1}$

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,

$S_2-S_1=1.70203_1-\dfrac{8.314}{28.97}\ln \dfrac{10\text { bar}}{1\text { bar}}\\S_2-S_1=2.36\rm\; kJ/kg K$

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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