Calculate the PV type of work for the followi | vedclass

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(C) The work done in a $PV$ system is given by the formula $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.Initial volume $V_1 = 150 \ mL + 150 \ mL = 3

Vedclass · Accepted Answer

$15.00$

Vedclass · Answer

(C) The work done in a $PV$ system is given by the formula $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.Initial volume $V_1 = 150 \ mL + 150 \ mL = 300 \ mL = 0.3 \ dm^3$.Final volume $V_2 = 150 \ mL = 0.15 \ dm^3$.External pressure $P_{ext} = 1 \ bar$.Substituting the values: $W = -1 \ bar 	imes (0.15 \ dm^3 - 0.3 \ dm^3) = -1 	imes (-0.15) \ dm^3 \ bar = 0.15 \ dm^3 \ bar$.Since $1 \ dm^3 \ bar = 100 \ J$,the work done is $0.15 	imes 100 \ J = 15.00 \ J$.

Calculate the $PV$ type of work for the following reaction at $1 \ bar$ pressure. $C_3H_{6(g)} (150 \ mL) + HCl_{(g)} (150 \ mL) \longrightarrow C_3H_7Cl_{(g)} (150 \ mL)$ (in $J$)

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