Question

Consider an isolated system and process as shown in the figure. (Process is that Seeta sitting on the freely movable piston jumps out). Assume no effect of reaction force due to jump on the system. Pressure equivalent of piston is 9 atm and pressure equivalent due to Seeta on the piston is 10 atm. Atmospheric pressure is 1 atm. (given molar heat capacity at constant volume $C_V=3Calmo{l}^{-1}{K}^{-1}$ )

The change in entropy of the system going from state-1 to state-2 is:
(ln 2 = 0.7, ln 3 = 1.1, ln 5 = 1.6, log 2 = 0.3)

• A. 0 cal/K
• B. 2.303 Cal/K
• C. 14.9 Cal/K
• D. None of these

Answer 1

The correct option is B 2.303 Cal/K

Entropy change for the system is given by,

$\Delta S_{sys}=n{C}_{v}ln\frac{T_2}{T_1}+nRln\frac{v_2}{v_1}$

q=u-w

$\int \frac{q_{rev}}{T}=\int \frac{n{C}_{v}dTUIKeyInputDownArrow}{T}+\int \frac{pdV}{T}$

$=10\times 3ln\frac{4}{5}+10\times 2ln\frac{8}{5}$ $[\frac{v_2}{v_1}=\frac{T_2}{T_1}\frac{P_1}{P_2}=\frac{4}{5}\times \frac{2}{1}=\frac{8}{5}]$

$=10\times 3ln\frac{2^2}{5}+10\times 2ln\frac{2^3}{5}$

$=60ln2-30ln5+60ln2-20ln5$

$=10(6ln2-3ln5+6ln2-2ln5)$

$=10(12ln2-5ln5)$

$=12\times 0.7-2\times 1.6$

$=(8.4-3.2)\times 10=52$

$\Delta S_{sys}=10\times 2.303[17log2-5]$

$=10\times 2.303[5.1-5]$

$=2.303Cal/K$