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Question

(C) The work done against constant external pressure is given by $W = -P_{ext} \Delta V$. Given: $P_{ext} = 1 \, atm$,$V_1 = 2 \, L$,$V_2 = 6 \, L$. $

Vedclass · Accepted Answer

$+396$

Vedclass · Answer

(C) The work done against constant external pressure is given by $W = -P_{ext} \Delta V$. Given: $P_{ext} = 1 \, atm$,$V_1 = 2 \, L$,$V_2 = 6 \, L$. $\Delta V = 6 \, L - 2 \, L = 4 \, L$. $W = -1 \, atm 	imes 4 \, L = -4 \, atm \cdot L$. Since $1 \, atm \cdot L = 101.325 \, J$,then $W = -4 	imes 101.325 \, J = -405.3 \, J$. According to the first law of thermodynamics,$\Delta U = q + W$. Given $q = +800 \, J$ (heat absorbed). $\Delta U = 800 \, J - 405.3 \, J = 394.7 \, J$. Rounding to the nearest provided option,$\Delta U \approx 396 \, J$.

$A$ gas present in a cylinder expands against a constant pressure of $1 \, atm$ from a volume of $2 \, L$ to a volume of $6 \, L$. In doing so,it absorbs $800 \, J$ of heat from the surroundings. The change in internal energy of the process is ....... $J$.

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