Calculate the amount of work done during isot | vedclass

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(C) Given: $V_{1} = 4 \ dm^{3}$,$V_{2} = 6 \ dm^{3}$,$P_{ex} = 3 \ atm$.The formula for work done against constant external pressure is $W = -P_{ex} \

Vedclass · Accepted Answer

$-607.8$

Vedclass · Answer

(C) Given: $V_{1} = 4 \ dm^{3}$,$V_{2} = 6 \ dm^{3}$,$P_{ex} = 3 \ atm$.The formula for work done against constant external pressure is $W = -P_{ex} \Delta V$.$\Delta V = V_{2} - V_{1} = 6 \ dm^{3} - 4 \ dm^{3} = 2 \ dm^{3} = 2 \ L$.$W = -3 \ atm 	imes 2 \ L = -6 \ L \cdot atm$.Since $1 \ L \cdot atm = 101.3 \ J$,we have $W = -6 	imes 101.3 \ J = -607.8 \ J$.

Calculate the amount of work done during isothermal expansion of a gas from a volume of $4 \ dm^{3}$ to $6 \ dm^{3}$ against a constant external pressure of $3 \ atm$ (in $J$)?

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