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(A) The formula for work done during expansion against constant external pressure is $W = -P_{ext} \cdot \Delta V$.Given $W = -6 \ dm^3 \ bar$ and $\D

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$1.2 	imes 10^5 \ Pa$

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(A) The formula for work done during expansion against constant external pressure is $W = -P_{ext} \cdot \Delta V$.Given $W = -6 \ dm^3 \ bar$ and $\Delta V = V_2 - V_1 = (20 - 15) \ dm^3 = 5 \ dm^3$.Substituting the values: $-6 \ dm^3 \ bar = -P_{ext} \cdot (5 \ dm^3)$.$P_{ext} = \frac{6}{5} \ bar = 1.2 \ bar$.Since $1 \ bar = 10^5 \ Pa$,the external pressure is $1.2 	imes 10^5 \ Pa$.

Two moles of an ideal gas are expanded isothermally from $15 \ dm^3$ to $20 \ dm^3$. If the amount of work done is $-6 \ dm^3 \ bar$,find the external pressure needed to obtain this work.

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