An ideal gas undergoes a process from state 1 (T₁ = 300 K,p₁ = 100 kPa) to state 2 T (T₂ = 600 K,p₂ = 500 kPa) . The specific heats of the ideal gas are: c_p = 1 kJ/kg-K and C_v = 0.7 kJ/kg-K. The change in specific entropy of the ideal gas from state 1 to state 2 (in kJ/kg-K) is ________ (correct to two decimal places)

To find the change in specific entropy of the ideal gas during the process from state 1 to state 2, we can use the fundamental thermodynamic relation for the entropy change of an ideal gas.

The change in specific entropy ($s_2-s_1$) of an ideal gas in terms of temperature and pressure is given by the formula:
$s_2-s_1=c_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{p_2}{p_1}\right)$
where:
• $c_p$ is the specific heat capacity at constant pressure.
• $R$ is the characteristic gas constant of the ideal gas.
• $T_1$ and $T_2$ are the initial and final absolute temperatures, respectively.
• $p_1$ and $p_2$ are the initial and final pressures, respectively.

Step 1: Calculate the gas constant ($R$)
According to Mayer's relation for an ideal gas:
$R=c_p-c_v$
Given:
• $c_p=1\text{kJ/kg-K}$
• $c_v=0.7\text{kJ/kg-K}$
Substituting these values:
$R=1-0.7=0.3\text{kJ/kg-K}$

Step 2: Substitute the state parameters into the entropy change equation
The given state parameters are:
• State 1: $T_1=300\text{K}$, $p_1=100\text{kPa}$
• State 2: $T_2=600\text{K}$, $p_2=500\text{kPa}$
Substituting the values of $c_p$, $R$, temperatures, and pressures:
$s_2-s_1=1\times \ln \left(\frac{600}{300}\right)-0.3\times \ln \left(\frac{500}{100}\right)$
Simplify the logarithmic terms:
$s_2-s_1=\ln \left(2\right)-0.3\times \ln \left(5\right)$

Step 3: Perform the final calculation
Using the natural logarithm values:
• $\ln \left(2\right)\approx 0.69315$
• $\ln \left(5\right)\approx 1.60944$
Substitute these approximations:
$s_2-s_1\approx 0.69315-0.3\times 1.60944$
$s_2-s_1\approx 0.69315-0.48283$
$s_2-s_1\approx 0.21032\text{kJ/kg-K}$

Rounding to two decimal places, the change in specific entropy is 0.21 kJ/kg-K.