Question

Common Data Question: Nitrogen gas (molecular weight $28$) is enclosed in a cylinder by a piston, at the initial condition of $2$ bar, $298$ K and $1m^3$. In a particular process, the gas slowly expands under isothermal condition, until the volume becomes $2m^3$. Heat exchange occurs with the atmosphere at $298K$ during this process.

The entropy changs for the universe during the process in $kJ/K$ is

• A. $0$
• B. $-0.6711$
• C. $0.4652$
• D. $0.0067$

Answer 1

The correct option is A $0$

Given process is isothermal, so change in entropy of universe

$(dS{)}_{universe}=(dS{)}_{system}+(dS{)}_{surrounding}$
Work done in isothermal process
$=P_1{V}_{1}In\left(V_2/V_1\right)$
$=200times1In\left(2/1\right)$
$=138.62kJ$
$\delta Q=dU+\delta W$
$\delta Q=138.62kJ\because dU=0)$
Heat taken by surroundings to the system
$=138.62kJ$
Change in entropy of surrounding
$(dS{)}_{suff}==\frac{-138.62}{298}=-0.465kJ/K$
Mass of nitrgen gas $=\frac{-132.62}{298}=-0.465kJ/K$
Mass of nitrogen gas $=\frac{PV}{RT}=\frac{200\times 1}{\frac{8.314}{28}\times 298}$
or $m=2.26kg$
Change in entropy of system
$\left(\frac{T_2}{T_1}\right)$
$mRin\left(\frac{V_2}{V_2}\right)$
$=2.26\times \frac{8.314}{28}In\left(\frac{2}{1}\right)$
$=2.26\times \frac{8.314}{28}In\left(\frac{2}{1}\right)$
$+0.465kJ/K$
$(dS{)}_{universe}=0.465-0.465$
$=0$