Pressure of Ideal gas

Classical Mechanics Level 3

The above left is a cylinder containing a given mass of ideal gas placed on a thermal reservoir with a constant temperature. The cylinder is separated by a divider into A and B of the same volume. Hence, the volumes of the ideal gases are the same as $V.$ The masses of the ideal gases in A and B are the same, and the pressures in A and B are also the same, which is $P$ . Now, as in the above right, we move the divider to the right until the volume of B is reduced by half. What is the resulting difference in pressure $P_B-P_A?$

\frac{3}{2}

\frac{2}{3}

\frac{4}{3}

\frac{1}{3}

2 solutions

Christopher Boo
Mar 8, 2014

According to Boyle's Law , as the temperature does not change, the relationship between the volume and pressure before and after can be expressed as:

$pV=p_AV_A$ and $pV=p_BV_B$

Given that,

$V_A=\frac{3}{2}V$ and $V_B=\frac{1}{2}V$

Then

$p_A=\frac{2}{3}p$ and $p_B=2p$

Hence,

$p_B-p_A$

$=2p-\frac{2}{3}p$

$=\frac{4}{3}p$

I dissent once you has put a divider you have to separated systems and isolated them each one will have a 1/2 of initial volume V. Now you must expand and compress them individually. And precisely on this resulting volume is our initial data for calculations.

Suppose you fix the divider and move each of the other wall of cylinder if we expand just one up to get double volume to get pressure in this size have of the initial pressure once it was divided. Doing the opposite in the other size you get double pressure so the answer will be 2-1/2 equal 3/2.

Mariano PerezdelaCruz - 6 years, 2 months ago

Gp 9210
Mar 6, 2014

use PV=NRT , with usual meaning

Pressure of Ideal gas

2 solutions

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