Question

The temperature (T) of one mole of an ideal gas varies with its volume (V) as $T=-\alpha {{V}^{3}}+\beta {{V}^{2}}$, where α and β are positive constants. The maximum pressure of the gas during this process is:
A) $\dfrac{\alpha \beta }{2R}$
B) $\dfrac{{{\beta }^{2}}R}{4\alpha }$
C) $\dfrac{\left( \alpha +\beta \right)R}{2{{\beta }^{2}}}$
D) $\dfrac{{{\alpha }^{2}}R}{2\beta }$

Answer 1

Hint: This problem can be solved by using the ideal gas equation to find a relation between the pressure P and the volume V of the gas. Then we can use the first and second order derivatives to find out the maxima of the pressure.
For maxima, the first derivative of the pressure with respect to volume should be zero, and the second derivative at that volume value should be negative.
Formula used: For a gas with pressure P, Volume V, molar amount n and temperature T, by the ideal gas equation,
$PV=nRT$
where R is the universal gas constant = $8.314J.mo{{l}^{-1}}.{{K}^{-1}}$.
For a plot of quantity x as a function of some other quantity t, for the maximum value of x,
$\dfrac{dx}{dt}=0$
and, $\dfrac{{{d}^{2}}x}{d{{t}^{2}}}<0$ for t at which $\dfrac{dx}{dt}=0$.

Complete step by step answer:
Firstly we will analyse the information given to us.
We are given an ideal gas.
Let the pressure of the gas be P.
Volume of the gas be V.
No. of moles of the gas (n) = 1. --(1)
Temperature of the gas is T.
We are also given that the temperature varies with the volume as
$T=-\alpha {{V}^{3}}+\beta {{V}^{2}}$ where α and β are positive constants. --(2)
Now, by the ideal gas equation,
$PV=nRT$ --(3)
where R is the universal gas constant = $8.314J.mo{{l}^{-1}}.{{K}^{-1}}$.
Now, using (1) and (2) in (3), we get,
$PV=R\left( -\alpha {{V}^{3}}+\beta {{V}^{2}} \right)$
$\therefore P=R\left( -\alpha {{V}^{2}}+\beta V \right)$ --(4)
For a plot of quantity x as a function of some other quantity t, for the maximum value of x,
$\dfrac{dx}{dt}=0$
and, $\dfrac{{{d}^{2}}x}{d{{t}^{2}}}<0$ for t at which $\dfrac{dx}{dt}=0$.
Now, to find out the maximum value of P, we will differentiate P with respect to V.
Differentiating equation (4) with respect to V, we get,
$\dfrac{dP}{dV}=R\left( -2\alpha V+\beta \right)$ --(5)
$\dfrac{{{d}^{2}}P}{d{{V}^{2}}}=-2\alpha R$ --(6)
For, maximum eq (5) = 0 as explained above.
$R\left( -2\alpha V+\beta \right)=0$
$\therefore V=\dfrac{\beta }{2\alpha }$ ---(7)
For maximum, equation (6) must be less than zero as explained above. This is true, since $-2\alpha R$ is negative as R (universal gas constant) and α are positive constant.
Hence, this condition is also satisfied.
Putting (7) in (4) we get,
$P=R\left( -\alpha \dfrac{{{\beta }^{2}}}{4{{\alpha }^{2}}}+\beta \dfrac{\beta }{2\alpha } \right)=R\left( \dfrac{-{{\beta }^{2}}}{4\alpha }+\dfrac{{{\beta }^{2}}}{2\alpha } \right)=\dfrac{{{\beta }^{2}}R}{4\alpha }$
Hence, the maximum value of pressure in this process will be $\dfrac{{{\beta }^{2}}R}{4\alpha }$.
Hence, the correct option is B) $\dfrac{{{\beta }^{2}}R}{4\alpha }$.

Note: The ideal gas equation comes in handy in all such problems. It is the most fundamental and important equation of this topic.
Sometimes, students may arrive at two answers for the first derivative being zero, these can be for the maximum and minimum respectively. Thus, it is always a good practice to do the second derivative test also and to check whether it is negative (for maxima) or positive (for minima).
Jumping to conclusions after only doing the first derivative test can lead to a silly mistake.