«

Irodov Solutions (2.7-2.9)

This post contains the solutions to the problems 2.7, 2.8 and 2.9 of I.E. Irodov’s “Problems in General Physics”. If you don’t really care about physics, you probably want to ignore this post.

2.7

Let the initial pressure be p_{o}=\gamma . After the first stroke, the pressure is given by

p_{1}=p_{0}\left(1-\frac{\Delta V}{V}\right)=\gamma\left(1-\frac{\Delta V}{V}\right)

After the second stroke, the pressure is given by

p_{2}=p_{1}\left(1-\frac{\Delta V}{V}\right)=\gamma\left(1-\frac{\Delta V}{V}\right)^{2}

Generalizing, the pressure in the vessel after the nth stroke will be given by

p_{n}=\gamma\left(1-\frac{\Delta V}{V}\right)^{n}

After the pressure is reduced \eta times, the pressure will be

p=\frac{\gamma}{\eta}

Assuming the pressure is reduced \eta times after x strokes, we get the equation

\frac{\gamma}{\eta}=\gamma\left(1-\frac{\Delta V}{V}\right)^{x}

Cancelling \gamma and applying a logarithm on both sides (since both sides of the equations must be positive), we get

-\ln\eta=x\ln\left(1-\frac{\Delta V}{V}\right)

Rearranging,

x=\frac{\ln\eta}{ln\frac{1}{\left(1-\frac{\Delta V}{V}\right)}}

Since \frac{\Delta V}{V} is small, \frac{1}{\left(1-\frac{\Delta V}{V}\right)}\thickapprox\left(1+\frac{\Delta V}{V}\right) . Hence, we get

x=\frac{\ln\eta}{\ln\left(1+\frac{\Delta V}{V}\right)}

This is the required expression.

2.8

Let the pressure at some time be p. After a differential time interval dt, the new pressure is given by

p_{f}=p\left(1-\frac{C}{V}\, dt\right)

Hence, the differential change in pressure is

dp=-p\,\frac{C}{V}\, dt

Separating variables and integrating both sides, we get

\intop_{p_{o}}^{p}\frac{dp}{p}=\intop_{0}^{t}-\frac{C}{V}\, dt

After doing the definite integration, we get an expression for the pressure as the required function of time.

p=p_{o}e^{-\frac{C}{V}t}

2.9

This can be solved directly using the expression we obtained while solving the previous question- p=p_{o}e^{-\frac{C}{V}t}.

\frac{p_{o}}{\eta}=p_{o}e^{-\frac{C}{V}t}

e^{-\frac{C}{V}t}=\frac{1}{\eta}

\frac{C}{V}t=\ln\eta

t=\frac{V}{C}\ln\eta

Substituting the values given in the question, we get the required answer, t=60s .

Tags: ,

This entry was posted on Monday, January 18th, 2010 at 5:21 pm and is filed under Physics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Leave a Reply