Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • During reversible adiabatic expansion of an ideal monoatomic gas, the final volume becomes 20 times to initial volume. The ratio (Final Temperature/Initial Temperature) will be equal to:<d

During reversible adiabatic expansion of an ideal monoatomic gas, the final volume becomes 20 times to initial volume. The ratio (Final Temperature/Initial Temperature) will be equal to:

Option: 1

(20)^{\frac{2}{3}}


Option: 2

(20)^{\frac{5}{3}}


Option: 3

(\frac{1}{20})^{\frac{5}{3}}


Option: 4

(\frac{1}{20})^{\frac{2}{3}}


Answers (1)

best_answer

For a reversible adiabatic change,

TV^{\gamma -1}= constant
Value of  \gamma  for a monoatomic gas is equal to \frac{5}{3}

\frac{T_{2}}{T_{1}}=\frac{V_{1}^{\gamma -1}}{V_{2}^{\gamma -1}}

\frac{T_{2}}{T_{1}}=(\frac{1}{20})^{\gamma -1}

\frac{T_{2}}{T_{1}}=(\frac{1}{20})^{\frac{2}{3}}

Hence, option number (4) is correct

Posted by

shivangi.shekhar

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

Punjab’s new 2-year MBBS bond, fee hike ‘exploitation’, ‘unfair’; interns, doctors shut OPDs in protest
anam-khan

NEET UG 2025 counselling likely to begin on July 1; AIQ seat allocation, court hearing updates
anam-khan

NEET UG: Delhi HC seeks NTA's response on plea alleging defective biometric system at centre

Latest Question