Air of mass 1 kg, initially at 300 K and 10 bar, is allowed to expand isothermally till it reaches a pressure of 1 bar. Assuming air as an ideal gas with gas constant of 0.287 kJ/kgK, the change in entropy of air (in kJ/kgK, round off to two decimal places) is ________.

The correct answer is 0.66.

Step-by-Step Explanation:

Given data:
Mass of air, m = 1 kg
Initial pressure, P₁ = 10 bar
Final pressure, P₂ = 1 bar
Initial temperature, T₁ = 300 K
Since the process is isothermal, the temperature remains constant:
T₂ = T₁ = 300 K
Gas constant of air, R = 0.287 kJ/kgK

The general expression for the change in entropy of an ideal gas in terms of temperature and pressure is given by:
$\Delta S=S_2-S_1=m{c}_p\ln \left(\frac{T_2}{T_1}\right)-mR\ln \left(\frac{P_2}{P_1}\right)$

For an isothermal process, T₁ = T₂. Therefore, $\ln \left(\frac{T_2}{T_1}\right)=\ln \left(1\right)=0$.

The equation simplifies to:
$\Delta S=-mR\ln \left(\frac{P_2}{P_1}\right)$

Substituting the values into the simplified equation:
$\Delta S=-\left(1\text{kg}\right)\times \left(0.287\text{kJ/kgK}\right)\times \ln \left(\frac{1}{10}\right)$

Using the property of logarithms, $\ln \left(\frac{1}{10}\right)=-\ln \left(10\right)$:
$\Delta S=0.287\times \ln \left(10\right)$
$\Delta S\approx 0.287\times 2.3026$
$\Delta S\approx 0.6608\text{kJ/kgK}$

Rounding off to two decimal places, the change in entropy of the air is 0.66 kJ/kgK.