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One mole of an ideal gas is allowed to expand from 1.0 L to 6.0 L at $27^{0} C$ against a constant pressure of 2.0 atm. The amount of work done during this process is ……(1 L atm = 101.3 J)

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2026 J

1013 J

506.5 J

300 J

answer is B.

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Detailed Solution

$w = P (V_{2} - V_{1}) = 2.0 (6 - 1) = 10 L a t m$

1 L atm = 101.3 J, 10 L atm = 1013 J .

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