The temperature (T) of one mole of an ideal g | vedclass

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(B) Given,$T = -\alpha V^3 + \beta V^2$ and the ideal gas equation for $n = 1$ mole is $PV = RT$,which implies $P = \frac{RT}{V}$.Substituting the exp

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$\frac{\beta^2 R}{4 \alpha}$

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(B) Given,$T = -\alpha V^3 + \beta V^2$ and the ideal gas equation for $n = 1$ mole is $PV = RT$,which implies $P = \frac{RT}{V}$.Substituting the expression for $T$ into the pressure equation:$P = \frac{R}{V} (-\alpha V^3 + \beta V^2) = R(-\alpha V^2 + \beta V)$.To find the maximum pressure,we differentiate $P$ with respect to $V$ and set it to zero:$\frac{dP}{dV} = R(-2\alpha V + \beta) = 0$.This gives $V = \frac{\beta}{2\alpha}$.To verify it is a maximum,we check the second derivative: $\frac{d^2P}{dV^2} = -2\alpha R$,which is negative since $\alpha > 0$,confirming a maximum.Substituting $V = \frac{\beta}{2\alpha}$ back into the expression for $P$:$P_{max} = R \left( -\alpha \left( \frac{\beta}{2\alpha} \right)^2 + \beta \left( \frac{\beta}{2\alpha} \right) \right) = R \left( -\frac{\beta^2}{4\alpha} + \frac{\beta^2}{2\alpha} \right) = R \left( \frac{\beta^2}{4\alpha} \right) = \frac{\beta^2 R}{4\alpha}$.

Column-$I$	Column-$II$
$(a)$ Kinetic energy per unit mole of gas.	$(i)$ $\frac{1}{2}RT$
$(b)$ Kinetic energy per one molecule of gas.	$(ii)$ $\frac{3}{2}RT$
	$(iii)$ $\frac{3}{2}k_BT$

The temperature $(T)$ of one mole of an ideal gas varies with its volume $(V)$ as $T = -\alpha V^3 + \beta V^2$,where $\alpha$ and $\beta$ are positive constants. The maximum pressure of the gas during this process is ............

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Column-$I$ represents physical quantity and Column-$II$ represents formula. Match them correctly:
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$(a)$ Kinetic energy per unit mole of gas. $(i)$ $\frac{1}{2}RT$
$(b)$ Kinetic energy per one molecule of gas. $(ii)$ $\frac{3}{2}RT$
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