Question

If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume by a formula of the form

Answer 1

Answer:

(a)  $\frac{dP}{dV}=\frac{2an^2}{V^3}-\frac{nRT}{(V-nb)^2}$

(b) $\frac{dP}{dt}=\frac{nR(V-nb)\frac{dT}{dt}-nRT\frac{dV}{dt}}{(V-nb)^2}+\frac{2an^2\frac{dV}{dt}}{V^3}$

(c) $\frac{dP}{dV}$ is the rate change of pressure of the gas per unit volume.

$\frac{dP}{dt}$ is the rate change of pressure of the gas per unit time.

Step-by-step explanation:

Formula:

• $\frac{d}{dx}(\frac uv)=\frac {v\frac{d}{dx}v-u\frac{d}{dx}v}{v^2}$

Given that,

$P=\frac{nRT}{V-nb}-\frac{an^2}{V^2}$

(a)

$P=\frac{nRT}{V-nb}-\frac{an^2}{V^2}$

Differentiating with respect to V

$\frac{dP}{dV}=\frac{(V-nb)\frac{d}{dV}(nRT)-(nRT)\frac{d}{dV}(V-nb)}{(V-nb)^2}-\frac{V^2\frac{d}{dV}(an^2)-(an^2)\frac{d}{dV}V^2}{(V^2)^2}$

      $=\frac {-nRT}{(V-nb)^2}-\frac{-an^2. 2V}{V^4}$

     $=\frac{2an^2}{V^3}-\frac{nRT}{(V-nb)^2}$

(b)

$P=\frac{nRT}{V-nb}-\frac{an^2}{V^2}$

Differentiating with respect to t

$\frac{dP}{dt}=\frac{(V-nb)\frac{d}{dt}(nRT)-(nRT)\frac{d}{dt}(V-nb)}{(V-nb)^2}-\frac{V^2\frac{d}{dt}(an^2)-(an^2)\frac{d}{dt}V^2}{(V^2)^2}$

    $=\frac{(V-nb)nR\frac{dT}{dt}-nRT\frac{dV}{dt}}{(V-nb)^2}-\frac{-an^2.2V\frac{dV}{dt}}{V^4}$

    $=\frac{nR(V-nb)\frac{dT}{dt}-nRT\frac{dV}{dt}}{(V-nb)^2}+\frac{2an^2\frac{dV}{dt}}{V^3}$

(c)

$\frac{dP}{dV}$ is the rate change of pressure of the gas per unit volume.

$\frac{dP}{dt}$ is the rate change of pressure of the gas per unit time.