Question

The temperature T of one mole of an ideal gas varies with its volume V as T=-αV³+βV², were α and β are positive constants. The maximum pressure of gas during this process is

Answer 1

it is given that Temperature T of one mole of an ideal gas varies with its volume as
$\bf{T=-\alpha V^3+\beta V^2}$

we know, from ideal gas equation PV = nRT
because no of mole of gas , n = 1
so, PV = RT , T = PV/R

now, $\frac{PV}{R}=-\alpha V^3+\beta V^2$

or, $P=\frac{R}{V}\left(-\alpha V^3+\beta V^2\right)$

or, $P=R\left(-\alpha V^2+\beta V\right)$ ......(1)

now differentiate P with respect to V,

dP/dV = $(-2\alpha V+\beta)R$

put dP/dV = 0, then, V = $\frac{\beta}{2\alpha}$

again differentiate with respect to V,

d²P/dV² = $-2\alpha\frac{\beta}{2\alpha}$

d²P/dV² = $-\beta R$ < 0 [ as β is positive constant.]

so, at V = $\frac{\beta}{2\alpha}$ P attains maximum value .

so, put value of V in equation (1),

we get, maximum value of P = $\frac{\beta^2}{4\alpha}R$

Answer 2

The answer is provided in the attachment.
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