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(A) For an adiabatic process, the relation between pressure and volume is given by $PV^{\gamma} = 	ext{constant}$, where $\gamma = \frac{C_P}{C_V}$.T

Vedclass · Accepted Answer

$\frac{4}{3}$

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(A) For an adiabatic process, the relation between pressure and volume is given by $PV^{\gamma} = 	ext{constant}$, where $\gamma = \frac{C_P}{C_V}$.Therefore, $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$, which implies $\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} ight)^{\gamma}$.Given, $V_1 = 8 \ L$, $V_2 = 27 \ L$, and $\frac{P_2}{P_1} = \frac{16}{81}$.Substituting these values, we get $\frac{16}{81} = \left( \frac{8}{27} ight)^{\gamma}$.We can write $\frac{16}{81}$ as $\left( \frac{2}{3} ight)^4$ and $\frac{8}{27}$ as $\left( \frac{2}{3} ight)^3$.So, $\left( \frac{2}{3} ight)^4 = \left( \left( \frac{2}{3} ight)^3 ight)^{\gamma} = \left( \frac{2}{3} ight)^{3\gamma}$.Comparing the exponents, we get $4 = 3\gamma$, which gives $\gamma = \frac{4}{3}$.

$A$ hypothetical gas expands adiabatically such that its volume changes from $8 \ L$ to $27 \ L$. If the ratio of final pressure of the gas to initial pressure of the gas is $\frac{16}{81}$, then the ratio of $\frac{C_P}{C_V}$ will be:

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