One mole He gas at 300 K, 10 atm is heated through a process in which its pressure is directly proportional to square of its absolute temperature till its temperature becomes 600 K The change in entropy of He is :

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

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

An Intiative by Sri Chaitanya

$\frac{5}{2} R \ln 2$

$\frac{1}{2} R \ln 2$

$\frac{3}{2} R \ln 2$

$\frac{9}{2} R In 2$

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

Detailed Solution

$\begin{array}{l} ∵ P \propto T^{2} \\ \frac{P_{1}}{P_{2}} = {(\frac{T_{1}}{T_{2}})}^{2} = \frac{1}{4} \\ Δ S = n C_{p} \ln (\frac{T_{2}}{T_{1}}) + n R \ln (\frac{P_{1}}{P_{2}}) \\ = 1 \times \frac{5}{2} R \ln 2 + 1 \times R \ln (\frac{1}{4}) \end{array}$

Watch 3-min video & get full concept clarity

hear from our champions

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET