Question

An adiabatic rigid chamber gas of 5 $m^3$ capacity at ${20}^{0}C$ and 220 kPa.1100

kJ of paddle shaft work is given to the chamber.

Take $C_p=1kJ/kgK,C_v=0.7kJ/kgK,R=0.30kJ/kgK$. What is the entropy increase of gas during the process?

• A. 0.250 kJ/K
• B. 3.780 kJ/k
• C. 3.124kJ/K
• D. 4.463kJ/K

Answer 1

The correct option is C 3.124kJ/K

(c) Correct Option
Since the chamber is adiabatic
$\delta Q=0$

From $I^{st}$ law of thermo dynamics

$\delta Q=\Delta u+\delta w$

$\Rightarrow \Delta U=-\delta W$

Now . $\delta W=-1100kJ$ (Work is done on the system)

$\Delta U=1100kJ$

$\Rightarrow m{c}_v(T_2-T_1)=1100kJ$

From ideal gas equation, before paddle work,

PV = mRT = m = PV/RT

$=\frac{220\times {10}^3\times 5}{300\times 293}\Rightarrow m=12.514kg$

$12.514\times 0.700(T_2-293)=1100$

$T_2=418.57K$

${}^{\left(\Delta S\right)}system=\underset{T_1}{\overset{T_2}{\int }}\frac{m{c}_{v}dT}{T}=m{c}_{v}ln\frac{T_2}{T_1}$

$=12.514\times 0.700\ln \left(\frac{418.57}{293}\right)=3.124kJ/K$