Question

One kilomole of an ideal gas is throttled from an initial pressure of $0.5$ MPa to $0.1$ MPa. The initial temperature is $300$ K. The entropy change of the universe is

• A. $0.0445$ kJ/K
• B. $4014.3$ kJ/K
• C. $13.38$ kJ/K
• D. $0.0446$ kJ/K

Answer 1

The correct option is C $13.38$ kJ/K

Given data: Number of mole
$n=1kmol$
$p_1=0.5MPa$
$p_2=0.1MPa$
$T_1=300K$
$=T_2$ for ideal gas
Entropy change of the universe,
$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}$
$=\Delta S_{sys}\because \Delta S_{surr}=0$
$=m{c}_{p}lo{g}_e\frac{T_2}{T_1}-mRlo{g}_e\frac{p_2}{p_1}$
$=-mRlo{g}_e\frac{p_2}{p_1}\because T_1=T_2$
(Throtting process)
$-m\frac{\overline{R}}{M}lo{g}_e\frac{p_2}{p_1}=-n\overline{R}lo{g}_e\frac{p_2}{p_1}$
$=-1\times 8.314lo{g}_e\frac{0.1}{0.5}$
$13.38kJ/K$
During throtting process,
Enthalpy change, $\Delta h=0$
$\because$ $h_2-h_1=0$
$h_1=h_2$
$c_P{T}_1=c_P{T}_2$ (for an ideal gas)
$T_1=T_2$
So, for an ideal gas (during throtting process) enthalpy is a function of temperature only and temperature
$T_1=T_2$