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NEET CHEMISTRYMedium

What is the amount of work done by an ideal gas, if the gas expands isothermally from 10−3 m310^{-3} \text{ m}^3 to 10−2 m310^{-2} \text{ m}^3 at 300 K300 \text{ K} against a constant pressure of 105 N m−210^5 \text{ N m}^{-2}?

A

+270 kJ+270 \text{ kJ}

B

−900 J-900 \text{ J}

C

+900 kJ+900 \text{ kJ}

D

−900 kJ-900 \text{ kJ}

Step-by-Step Solution

The work done during an isothermal expansion against a constant external pressure (pexp_{ex}) is given by the formula: W=−pexΔV=−pex(Vf−Vi)W = -p_{ex} \Delta V = -p_{ex} (V_f - V_i) Given: External pressure, pex=105 N m−2p_{ex} = 10^5 \text{ N m}^{-2} Initial volume, Vi=10−3 m3V_i = 10^{-3} \text{ m}^3 Final volume, Vf=10−2 m3V_f = 10^{-2} \text{ m}^3 Substituting the given values into the equation: W=−105 N m−2×(10−2−10−3) m3W = -10^5 \text{ N m}^{-2} \times (10^{-2} - 10^{-3}) \text{ m}^3 W=−105×(0.01−0.001) JW = -10^5 \times (0.01 - 0.001) \text{ J} W=−105×0.009 J=−900 JW = -10^5 \times 0.009 \text{ J} = -900 \text{ J}. Thus, the work done by the gas is −900 J-900 \text{ J}.

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