In a process,temperature and volume of one mo | vedclass

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Question

(A) Given the process relation: $VT = K$.Using the ideal gas law $PV = nRT$,we have $T = \frac{PV}{nR}$.Substituting $T$ into the process equation: $V

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$\frac{1}{2} R \Delta T$

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(A) Given the process relation: $VT = K$.Using the ideal gas law $PV = nRT$,we have $T = \frac{PV}{nR}$.Substituting $T$ into the process equation: $V \left( \frac{PV}{nR} ight) = K$.This simplifies to $PV^2 = nRK$,which is a polytropic process of the form $PV^x = 	ext{constant}$,where $x = 2$.For a monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.The molar heat capacity $C$ for a polytropic process is given by $C = C_v + \frac{R}{1-x}$.Substituting the values: $C = \frac{3}{2} R + \frac{R}{1-2} = \frac{3}{2} R - R = \frac{1}{2} R$.The heat absorbed by $n$ moles of gas is $\Delta Q = n C \Delta T$.Given $n = 1$,we get $\Delta Q = 1 	imes \left( \frac{1}{2} R ight) 	imes \Delta T = \frac{1}{2} R \Delta T$.

In a process,temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation $VT = K$,where $K$ is a constant. In this process,the temperature of the gas is increased by $\Delta T$. The amount of heat absorbed by the gas is ($R$ is the gas constant).

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