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Question

(A) The formula for work done during expansion against constant external pressure is $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.Given: $P_{ext} = 2

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$2 	imes 10^{-3} \ m^3$

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(A) The formula for work done during expansion against constant external pressure is $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.Given: $P_{ext} = 2.40 	imes 10^5 \ Pa$,$V_2 = 2.2 	imes 10^{-3} \ m^3$,$W = -0.048 \ kJ = -48 \ J$.Substituting the values: $-48 \ J = -2.4 	imes 10^5 \ Pa 	imes (2.2 	imes 10^{-3} \ m^3 - V_1)$.Dividing both sides by $-2.4 	imes 10^5 \ Pa$: $\frac{-48}{-2.4 	imes 10^5} = 2.2 	imes 10^{-3} - V_1$.$20 	imes 10^{-5} = 2.2 	imes 10^{-3} - V_1$.$0.2 	imes 10^{-3} = 2.2 	imes 10^{-3} - V_1$.$V_1 = 2.2 	imes 10^{-3} - 0.2 	imes 10^{-3} = 2.0 	imes 10^{-3} \ m^3$.

When a certain volume of gas expands against a constant external pressure of $2.40 \times 10^5 \ Pa$ at $300 \ K$ to a final volume of $2.2 \times 10^{-3} \ m^3$. If the work obtained is $-0.048 \ kJ$,what is the initial volume of the gas?

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