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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $P$:$P_{max} = R \left( -\alpha \left( \frac{\beta}{2\alpha} \right)^2 + \beta \left( \frac{\beta}{2\alpha} \right) \right) = R \left( -\frac{\beta^2}{4\alpha} + \frac{\beta^2}{2\alpha} \right) = R \left( \frac{\beta^2}{4\alpha} \right) = \frac{\beta^2 R}{4\alpha}$.

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 ............

Similar Questions

$A$ thermally insulated vessel contains an ideal gas of molecular mass $M$ and ratio of specific heats $\gamma$. It is moving with speed $v$ and it is suddenly brought to rest. Assuming no heat is lost to the surroundings,its temperature increases by

If the pressure of $CO_2$ (real gas) in a container is given by $P = \frac{RT}{2V - b} - \frac{a}{4V^2}$,then the mass of the gas in the container is ...... $g$.

If $v_1$ is the speed of sound in a diatomic gas at $273^{\circ}C$ and $v_2$ is the r.m.s. speed of its molecules at $273 \ K$,then $\frac{v_1}{v_2}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module