The volume of 1 ; mole of an ideal gas with t | vedclass

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Question

(B) Given the relation $V = \frac{b}{T}$,we can write $VT = b$ (constant).Using the ideal gas law $PV = nRT$,for $n = 1$ mole,we have $T = \frac{PV}{R

Vedclass · Accepted Answer

$\left( \frac{2 - \gamma}{\gamma - 1} \right) R \Delta T$

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(B) Given the relation $V = \frac{b}{T}$,we can write $VT = b$ (constant).Using the ideal gas law $PV = nRT$,for $n = 1$ mole,we have $T = \frac{PV}{R}$.Substituting this into the relation: $V \left( \frac{PV}{R} ight) = b$,which simplifies to $PV^2 = bR = 	ext{constant}$.This is a polytropic process of the form $PV^x = 	ext{constant}$,where $x = 2$.The molar heat capacity $C$ for a polytropic process is given by $C = \frac{R}{\gamma - 1} + \frac{R}{1 - x}$.Substituting $x = 2$: $C = \frac{R}{\gamma - 1} + \frac{R}{1 - 2} = \frac{R}{\gamma - 1} - R = R \left( \frac{1 - (\gamma - 1)}{\gamma - 1} ight) = R \left( \frac{2 - \gamma}{\gamma - 1} ight)$.The heat absorbed $Q$ is given by $Q = nC \Delta T$.For $n = 1$,$Q = \left( \frac{2 - \gamma}{\gamma - 1} ight) R \Delta T$.

The volume of $1 \; mole$ of an ideal gas with the adiabatic exponent $\gamma$ is changed according to the relation $V = \frac{b}{T}$,where $b$ is a constant. The amount of heat absorbed by the gas in the process if the temperature is increased by $\Delta T$ will be:

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