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

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(B) For an adiabatic process, the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.Given the relation $TV^{x} = \text{constant}$, 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 $x$:$x = \frac{5}{3} - 1 = \frac{2}{3}$.

Class 11 Physics Thermodynamics Adiabatic Process

Medium

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

A
$\frac{2}{5}$
B
$\frac{2}{3}$
C
$\frac{5}{3}$
D
$\frac{3}{5}$

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Thermodynamics

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Adiabatic Process

$A$ monoatomic gas $(\gamma = 5/3)$ at a temperature of $27^{\circ}C$ is compressed adiabatically to $8/27$ of its initial volume. The change in temperature is . . . . . . $K$.

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The volume of a gas is reduced adiabatically to $(1/4)^{th}$ of its initial volume. If the initial temperature is $27\,^{\circ}C$ and $\gamma = 1.4$, the new temperature is:

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$A$ gas at normal temperature is suddenly compressed to one-fourth of its original volume. If $\gamma = 1.5$, then the increase in the temperature of the gas in Kelvin is ($\gamma$ is the ratio of specific heats).

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$A$ 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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$A$ monoatomic gas $(\gamma = 5/3)$ at a temperature of $27^{\circ}C$ is compressed adiabatically to $8/27$ of its initial volume. The change in temperature is . . . . . . $K$.

$A$ sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is (given,$\gamma = 3/2$):

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