Question

How much work must be done on a system to decrease its volume from 20.0 l to 13.0 l by exerting a constant pressure of 4.0 atm?

Answer 1

By exerting the given constant pressure, work done on the system to decrease from initial volume to final volume is 2.84kJ.

Given the data in the question

• Initial volume; $v_1 = 20.0L$

• Final volume; $v_2 = 13.0L$

• Pressure exerted; $P = 4.0 atm$

The magnitude of the work done when a gas expands is equal to the product of the pressure of the gas and the change in the volume of the gas.

For work done by gas against a constant external pressure, we use the

expression:

$Work\ done = -P\ *\ \delta V$

We substitute our given values into the equation

$Work\ done = -P\ *\ ( v_2 - v_1)\\\\Work\ done = -4.0atm\ *\ ( 13.0L - 20.0L)\\\\Work\ done = -4.0atm\ *\ -7L\\\\Work\ done = 28L.atm\\\\Work\ done = 28L.atm\ *\ [ \frac{101.3J}{L.atm}] \\\\Work\ done = 2836.4J\\\\Work\ done = 2.84 kJ$

Therefore, by exerting the given constant pressure, work done on the system to decrease from initial volume to final volume is 2.84kJ.

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