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NEET CHEMISTRYMedium

The work done during the expansion of a gas from a volume of 4 dm34\text{ dm}^3 to 6 dm36\text{ dm}^3 against a constant external pressure of 3 atm3\text{ atm} is:

A

−608 J-608\text{ J}

B

+304 J+304\text{ J}

C

−304 J-304\text{ J}

D

−6.00 J-6.00\text{ J}

Step-by-Step Solution

The work done (ww) during the expansion of a gas against a constant external pressure (pextp_{ext}) is given by the formula: w=−pextΔV=−pext(Vf−Vi)w = -p_{ext} \Delta V = -p_{ext}(V_f - V_i) Given: Initial volume, Vi=4 dm3=4 LV_i = 4\text{ dm}^3 = 4\text{ L} (since 1 dm3=1 L1\text{ dm}^3 = 1\text{ L}) Final volume, Vf=6 dm3=6 LV_f = 6\text{ dm}^3 = 6\text{ L} External pressure, pext=3 atmp_{ext} = 3\text{ atm} Substituting the values: w=−3 atm×(6 L−4 L)=−3×2=−6 L atmw = -3\text{ atm} \times (6\text{ L} - 4\text{ L}) = -3 \times 2 = -6\text{ L atm} To convert the work done into Joules, we use the conversion factor 1 L atm=101.325 J1\text{ L atm} = 101.325\text{ J}: w=−6 L atm×101.325 J/L atm=−607.95 J≈−608 Jw = -6\text{ L atm} \times 101.325\text{ J/L atm} = -607.95\text{ J} \approx -608\text{ J}.

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