A monoatomic ideal gas undergoes an adiabatic | vedclass

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(B) For an adiabatic process, the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.Given the relation $TV

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$\frac{2}{3}$

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(B) For an adiabatic process, the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.Given the relation $TV^{x} = 	ext{constant}$, we compare the exponents of $V$.Thus, $x = \gamma - 1$.For a monoatomic ideal gas, the adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3}$.Substituting the value of $\gamma$ into the expression for $x$:$x = \frac{5}{3} - 1 = \frac{2}{3}$.

$A$ monoatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume is $TV^{x} = \text{constant}$, then $x$ is

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