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Work Done in Isothermal Reversible Process

What is the w...

Question

What is the work done by $0.2$ mole of a gas at room temperature to double volume during isobaric process (take $R = 2 c a l m o l^{- 1} K^{- 1}$ )

30

cal

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40

cal

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120

cal

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160

cal

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Solution

The correct option is C $120$ cal
Work done $W = P (V_{f} - V_{i}) = P (2 V - V) = P V$
So, we know that the work done $W = P V = n R T$
$W = 0.2 \times 2 \times 300 = 120 c a l$
Where $n$ is the mole of gas, $T$ is the temprature,

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