A monoatomic gas is taken through a process T | vedclass

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(C) Given the process equation: $TP^{-1/3} = 	ext{constant}$.Using the ideal gas law $PV = nRT$, we have $T \propto PV$, so $PV \cdot P^{-1/3} = 	ex

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Its temperature will decrease

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(C) Given the process equation: $TP^{-1/3} = 	ext{constant}$.Using the ideal gas law $PV = nRT$, we have $T \propto PV$, so $PV \cdot P^{-1/3} = 	ext{constant}$, which simplifies to $P^{2/3}V = 	ext{constant}$.Raising both sides to the power of $3/2$, we get $PV^{3/2} = 	ext{constant}$.This is a polytropic process $PV^x = 	ext{constant}$ where $x = 3/2$.For a monoatomic gas, the adiabatic index $\gamma = 5/3$.The molar heat capacity $C$ for a polytropic process is given by $C = \frac{R}{\gamma - 1} + \frac{R}{1 - x}$.Substituting the values: $C = \frac{R}{5/3 - 1} + \frac{R}{1 - 3/2} = \frac{R}{2/3} + \frac{R}{-1/2} = \frac{3R}{2} - 2R = -\frac{R}{2}$.Since the molar heat capacity $C$ is negative, from the relation $\Delta Q = nC\Delta T$, if heat is given to the gas $(\Delta Q > 0)$, then $\Delta T$ must be negative $(\Delta T Therefore, the temperature of the gas will decrease.

$A$ monoatomic gas is taken through a process $TP^{-1/3} = \text{constant}$. If heat is given to the gas, what happens to its temperature?

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