Question

A gas is compressed at constant temperature from a volume of 5.68 L to a volume of 2.35 L by an external pressure of 732 torr. Calculate the work done in J

Answer 1

Answer: The work done in J is 324

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

$W=-P\Delta V=-P(V_2-V_1)$

W = amount of work done = ?

P = pressure = 732 torr = 0.96 atm    (760torr =1atm)

$V_1$ = initial volume = 5.68 L

$V_2$ = final volume = 2.35  L

Putting values in above equation, we get:

$W=-0.96atm\times (2.35-5.68)L=3.20L.atm$

To convert this into joules, we use the conversion factor:

$1L.atm=101.33J$

So, $3.20L.atm=3.20\times 101.3=324J$

The positive sign indicates the work is done on the system

Hence, the work done for the given process is 324 J

Answer 2

The work done will be "324 J".

Given values:

• Pressure, $P = 732 \ torr \ or \ 0.96 \ atm$
• Initial volume, $V_1 = 5.68 \ L$
• Final volume, $V_2 = 2.35 \ L$

We know the equation,

→ $W = - P\Delta V$

       $= -P(V_2-V_1)$

By substituting the values,

       $= -0.96\times (2.35-5.68)$

       $= 3.20 \ L.atm$

By converting it in "J",

       $= 3.20\times 101.3$

       $= 324 \ J$

Thus the above answer is right.

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