Answer to Question #131235 in Classical Mechanics for giannis

Question

Hi i am a high school student! my question is
we have a gas and we dont care about temperature change. the pressure is given from the type P(V)=6/V and the starting V is Vo. what will happen to the pressure if V is increased by 8%.
I know from the theory that it should decrease by 8% due to boyle's law BUT my maths fail me and i cannot prove it

Answer 1

Hi!

Just write the law:

$$P=\frac{6}{V}.$$

You said the pressure is increased by 8% compared to the initial pressure $P_0$, which means

$$P=\frac{(100\%+8\%)}{100\%}P_0=1.08P_0.$$

What will happen to the volume? From the very first equation express volume by multiplying both sides of the equation by V and dividing both sides by P:

$$V=\frac{6}{P}.$$

Substitute our new pressure for P:

$$V=\frac{6}{1.08P_0}=\frac{5.56}{P_0}.$$

Now compare how is this different from initial volume, which is just 6/P?

$$\frac{V}{V_0}=\frac{5.56/P_0}{6/P_0}=0.926.$$

In other words (multiply both sides if this equation by V₀):

$$V=0.926V_0.$$

So, by how many percent is V smaller than V₀? Let's calculate this.

Find the difference between the initial and final pressure ($\Delta V=V_0-V$) and divide it by the initial value. Then multiply by 100% to express the answer in %:

$$\epsilon=\frac{\Delta V}{V_0}\cdot100\%=\frac{V_0-V}{V_0}\cdot100\%,$$

now it's time to substitute 0.926V₀ for V:

$$\epsilon=\frac{V_0-0.926V_0}{V_0}\cdot100\%=7.4\%.$$

Not 8%, but quite close.