An ideal gas is allowed to expand from 2 dm^ | vedclass

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(B) The formula for work done against constant external pressure is $W = -P_{ext} \Delta V$. First,convert the volumes to the same unit $(m^3)$: $V_1

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$-0.4$

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(B) The formula for work done against constant external pressure is $W = -P_{ext} \Delta V$. First,convert the volumes to the same unit $(m^3)$: $V_1 = 2 \ dm^3 = 2 	imes 10^{-3} \ m^3$. $V_2 = 6 	imes 10^{-3} \ m^3$. Change in volume $\Delta V = V_2 - V_1 = (6 - 2) 	imes 10^{-3} \ m^3 = 4 	imes 10^{-3} \ m^3$. Given external pressure $P_{ext} = 1 \ bar = 10^5 \ Pa = 10^5 \ N/m^2$. Substituting the values: $W = -(10^5 \ N/m^2) 	imes (4 	imes 10^{-3} \ m^3) = -400 \ J$. Converting to $kJ$: $W = -400 / 1000 \ kJ = -0.4 \ kJ$.

An ideal gas is allowed to expand from $2 \ dm^3$ to $6 \times 10^{-3} \ m^3$ against a constant external pressure of $1 \ bar$. The work done in $kJ$ is (in $kJ$)

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