An ideal gas at 27oC is compressed adiabatically to 827 of its original volume. If γ=53, then the rise in temperature is

T2T1=V1V2γ−1⇒T2=30027853−1=30027823=3002781/32=800322=675K⇒ΔT=675−300=375K

An ideal gas at 27oC is compressed adiabatically to 827 of its original volume. If γ=53, then the rise in temperature is

T2T1=V1V2γ−1⇒T2=30027853−1=30027823=3002781/32=800322=675K⇒ΔT=675−300=375K

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

An ideal gas at $27^{o} C$ is compressed adiabatically to $\frac{8}{27}$ of its original volume. If $γ = \frac{5}{3}$ , then the rise in temperature is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

405 K

225 K

450 K

375 K

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} \Rightarrow T_{2} = 300 {(\frac{27}{8})}^{\frac{5}{3} - 1} = 300 {(\frac{27}{8})}^{\frac{2}{3}} \\ = 300 {\{{(\frac{27}{8})}^{1 / 3}\}}^{2} = 800 {(\frac{3}{2})}^{2} = 675 K \\ \Rightarrow ΔT = 675 - 300 = 375 K \end{array}$