Two kg of water is converted into steam by bo | vedclass

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Question

(D) The work done by a system at constant pressure (isobaric process) is given by the formula $W = P \Delta V$.Here,the pressure $P$ is atmospheric pr

Vedclass · Accepted Answer

$340$

Vedclass · Answer

(D) The work done by a system at constant pressure (isobaric process) is given by the formula $W = P \Delta V$.Here,the pressure $P$ is atmospheric pressure,which is $1.013 	imes 10^5 \ Pa$.The change in volume $\Delta V = V_f - V_i = 3.34 \ m^3 - 2 	imes 10^{-3} \ m^3 = 3.34 - 0.002 = 3.338 \ m^3$.Substituting these values into the work formula:$W = (1.013 	imes 10^5 \ Pa) 	imes (3.338 \ m^3)$$W \approx 338,137 \ J$Converting to kilojoules $(kJ)$:$W \approx 338 \ kJ \approx 340 \ kJ$.Therefore,the work done by the system is approximately $340 \ kJ$.

Two kg of water is converted into steam by boiling at atmospheric pressure. The volume changes from $2 \times 10^{-3} \ m^3$ to $3.34 \ m^3$. The work done by the system is about ....... $kJ$.

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