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(A) The formula for work done during expansion against constant external pressure is $W = -P_{	ext{ext}} \Delta V$. Given: $W = -5.09 \ kJ = -5090 \

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$5.09 	imes 10^5 \ Nm^{-2}$

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(A) The formula for work done during expansion against constant external pressure is $W = -P_{	ext{ext}} \Delta V$. Given: $W = -5.09 \ kJ = -5090 \ J$,$V_1 = 2 	imes 10^{-2} \ m^3$,$V_2 = 3 	imes 10^{-2} \ m^3$. Change in volume $\Delta V = V_2 - V_1 = (3 - 2) 	imes 10^{-2} \ m^3 = 1 	imes 10^{-2} \ m^3$. Substituting the values: $-5090 = -P_{	ext{ext}} 	imes (1 	imes 10^{-2})$. $P_{	ext{ext}} = \frac{5090}{10^{-2}} = 509000 \ Nm^{-2} = 5.09 	imes 10^5 \ Nm^{-2}$.

What is the constant external pressure of an ideal gas when expanded from $2 \times 10^{-2} \ m^3$ to $3 \times 10^{-2} \ m^3$,if the work done by the gas is $-5.09 \ kJ$?

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