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The work don e dur ing the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm is (1L atm  = 101.32J) [2004]
  • a)
    – 6J
  • b)
    – 608J
  • c)
    + 304J
  • d)
    – 304J
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The work don e dur ing the expansion of a gas from a volume of 4 dm3 t...
W = – pΔV
= -3(6 - 4) = - 6 litre atmosphere
= -6 x 101.32 = -608J
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Most Upvoted Answer
The work don e dur ing the expansion of a gas from a volume of 4 dm3 t...
The work done during the expansion of a gas can be calculated using the formula:

Work = Pressure * Change in Volume

In this case, the pressure is constant at 3 atm and the change in volume is from 4 dm3 to 6 dm3.

Change in Volume = Final Volume - Initial Volume
= 6 dm3 - 4 dm3
= 2 dm3

Plugging these values into the formula, we get:

Work = 3 atm * 2 dm3

Since 1 dm3 = 1 L, we can convert dm3 to L:

Work = 3 atm * 2 L

Finally, we can convert atm L to J using the conversion factor 1 L atm = 101.32 J:

Work = 3 * 2 * 101.32 J
= 606.72 J

Therefore, the work done during the expansion of the gas is 606.72 J.
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