How many cycles until pressure decreases k times?

[Yegor]

Air pump is pumping out a gas from a vessel (vessel's volume is V).
After each cycle i it pumps out dV.
Gas is ideal. T=Const.
After how much cycles pressure P will decrease k times?

Here is my work.

PV = nRT
Initially n=n(0)

After each cycle n decreases (1-dV/V) times.
So, after x cycles n(x)=n(0)(1-dV/V)^x
k=p(0)/p(x)=n(0)/n(x)
n(x)/n(0)=(1-dV/V)^x
So x =ln(1/k)/ln(1-dV/V)

But the answer is x=ln(k)/ln(1+dV/V)

As I understand, they assume that after each cycle n decreases 1/(1+dV/V) times. In this way i really receive x=ln(k)/ln(1+dV/V)
I don't understand it. Help me please.

 

[Yegor]

Has anybody any ideas?

 

[Gamma]

They have done some mathematics with x =ln(1/k)/ln(1-dV/V) to arrive at a more nice looking answer.

x= -ln (k) / ln (1- dV/V)
= ln (k) / ln [(1-dV/V)] ^-1

now, [(1-dV/V)] ^-1 = 1 + dV/V (approximately)

so, x = ln (k) / ln [(1+dV/V)

 

[clive]

Each cycle corresponds to a T=const. process. (in another version of this problem the pumping process is very fast and a the transformation would correspond to an adiabatic process - all you have to do is to add the exponent $$\gamma$$ to the equations below). After the first cycle the air volume increases from V to V+dV and the pressure decreases at $$p_1$$. You have
$$p_0\cdot V=p_1\cdot (V+dV)$$
After the first cycle, the pump takes out the air from dV, and you get again
$$p_1=p_0 \frac{V}{V+dv}$$
In the second cycle,
$$p_1\cdot V=p_2\cdot (V+dV)$$
and
$$p_2=p_1 \frac{V}{V+dv}=p_0 \frac{V^2}{(V+dv)^2}$$

and so on...

$$p_n=p_{n-1} \frac{V}{V+dv}=p_0 \frac{V^n}{(V+dv)^n}$$