Question

41. One mole of a gas occupying 3dm3 expands
against a constant external pressure of 1 atm to
a volume of 13 lit. The workdone is -
PLEASE ANSWER WITH CORRECT EXPLAINATION​

Answer 1

Explanation:

Use formula of Mechanical Work:

W= Pext ( V2 - V1 )

where

• W - Mechanical work
• Pext - Constant External Pressure
• V2 - Final Volume (After Expansion or Compression of gas)
• V1 - Initial Volume is gas

Answer 2

The work done is -1013 Joules

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 = 1 atm

$V_1$ = initial volume = $3dm^3=3L$     $(1dm^3=1L)$

$V_2$ = final volume = 13 L

Putting values in above equation, we get:

$W=-1atm\times (13-3)L=-10L.atm$

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

$1L.atm=101.33J$

So, $-10L.atm=-10\times 101.3=-1013J$

The negative sign indicates the system is doing work.

Hence, the work done for the given process is -1013 J

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