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  • 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

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