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

Vedclass · Accepted Answer

$+395$

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(C) The work done against constant external pressure is given by the formula $W = -P_{ext} \Delta V$. Given: $P_{ext} = 1 \, atm$,$V_1 = 2 \, L$,$V_2 = 6 \, L$,and $q = +800 \, J$ (heat absorbed). First,calculate the change in volume: $\Delta V = V_2 - V_1 = 6 \, L - 2 \, L = 4 \, L$. Convert the work done from $L \cdot atm$ to $J$ using the conversion factor $1 \, L \cdot atm = 101.325 \, J$: $W = -1 \, atm 	imes 4 \, L = -4 \, L \cdot atm = -4 	imes 101.325 \, J = -405.3 \, J$. According to the first law of thermodynamics: $\Delta U = q + W$. Substituting the values: $\Delta U = 800 \, J + (-405.3 \, J) = 394.7 \, J$. Rounding to the nearest integer,we get $\Delta U \approx 395 \, J$.

$A$ 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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