An ideal gas expands from 1 times 10^{-3} m^ | vedclass

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Question

(A) The formula for work done during expansion against a constant external pressure is $W = -P_{ex} \Delta V$. Given: $V_{1} = 1 	imes 10^{-3} \ m^{3

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$-9 	imes 10^{2} \ J$

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(A) The formula for work done during expansion against a constant external pressure is $W = -P_{ex} \Delta V$. Given: $V_{1} = 1 	imes 10^{-3} \ m^{3} = 0.001 \ m^{3}$ $V_{2} = 1 	imes 10^{-2} \ m^{3} = 0.01 \ m^{3}$ $P_{ex} = 1 	imes 10^{5} \ Nm^{-2}$ Change in volume $\Delta V = V_{2} - V_{1} = 0.01 - 0.001 = 0.009 \ m^{3}$. Substituting the values: $W = -1 	imes 10^{5} 	imes 0.009 = -900 \ J = -9 	imes 10^{2} \ J$.

An ideal gas expands from $1 \times 10^{-3} \ m^{3}$ to $1 \times 10^{-2} \ m^{3}$ at $300 \ K$ against a constant external pressure of $1 \times 10^{5} \ Nm^{-2}$. The work done is:

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