One mole of an ideal diatomic gas Cv=5cal was transformed from initial 25°C and 1 L to the state when temperature is 100∘C and volume 10 L. The entropy change of the process can be expressed as : (R=2cal/mol/K)

7 ln ⁡ 373 298 + 2 ln ⁡ 1 10

5 ln ⁡ 373 298 + 2 ln ⁡ 1 10

Book Online Demo

Check Your IQ

Try Test

Courses

One mole of an ideal diatomic gas $(C_{v} = 5 cal)$ was transformed from initial $25 ° C$ and 1 L to the state when temperature is $100^{\circ} C$ and volume 10 L. The entropy change of the process can be expressed as : $(R = 2 cal / mol / K)$

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$5 \ln \frac{373}{298} + 2 \ln 10$

$3 \ln \frac{298}{373} + 2 \ln 10$

$7 \ln \frac{373}{298} + 2 \ln \frac{1}{10}$

$5 \ln \frac{373}{298} + 2 \ln \frac{1}{10}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Changes to the entropy formula in general

$Δ S_{s y n} = n C_{v} \ln (\frac{T_{2}}{T_{1}}) + n R \ln (\frac{V_{2}}{V_{1}})$

given value

$\begin{aligned} T_{1} = 25^{\circ} C = 298 k \end{aligned}$

$T_{2} = 100^{\circ} C = 373 K$

$C_{v} = 5 Cal$

$R = 2 Cal / molK$

$\begin{aligned} V_{1} = 1 L \\ V_{2} = 10 L \end{aligned}$

$ΔS = 5 \ln (\frac{373}{298}) + 2 \ln (\frac{10}{1})$

$= 5 \ln (\frac{373}{298}) + 2 \ln 10$

Watch 3-min video & get full concept clarity