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NEET CHEMISTRYMedium

Under isothermal condition, a gas at 300 K300 \text{ K} expands from 0.1 L0.1 \text{ L} to 0.25 L0.25 \text{ L} against a constant external pressure of 2 bar2 \text{ bar}. The work done by the gas is:

A

30 J30 \text{ J}

B

30 J-30 \text{ J}

C

5 kJ5 \text{ kJ}

D

25 J25 \text{ J}

Step-by-Step Solution

Work done during an irreversible expansion against a constant external pressure is given by the formula: W=pexΔV=pex(VfVi)W = -p_{ex}\Delta V = -p_{ex}(V_f - V_i)

Given: External pressure, pex=2 barp_{ex} = 2 \text{ bar} Initial volume, Vi=0.1 LV_i = 0.1 \text{ L} Final volume, Vf=0.25 LV_f = 0.25 \text{ L}

Substituting the values into the equation: W=2 bar×(0.25 L0.1 L)W = -2 \text{ bar} \times (0.25 \text{ L} - 0.1 \text{ L}) W=2 bar×0.15 L=0.30 L barW = -2 \text{ bar} \times 0.15 \text{ L} = -0.30 \text{ L bar}

To convert the work done into Joules, we use the conversion factor 1 L bar=100 J1 \text{ L bar} = 100 \text{ J}: W=0.30×100 J=30 JW = -0.30 \times 100 \text{ J} = -30 \text{ J}

Thus, the work done by the gas is 30 J-30 \text{ J}.

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