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(C) Given: $P = 1 \ atm$,$V_1 = 2 \ L$,$V_2 = 6 \ L$,$q = +800 \ J$ (heat absorbed).Change in volume $\Delta V = 6 - 2 = 4 \ L$.Work done $W = -P \Del

Vedclass · Accepted Answer

$394.95$

Vedclass · Answer

(C) Given: $P = 1 \ atm$,$V_1 = 2 \ L$,$V_2 = 6 \ L$,$q = +800 \ J$ (heat absorbed).Change in volume $\Delta V = 6 - 2 = 4 \ L$.Work done $W = -P \Delta V = -1 \ atm 	imes 4 \ L = -4 \ L \cdot atm$.Since $1 \ L \cdot atm = 101.325 \ J$,then $W = -4 	imes 101.325 = -405.3 \ J$.According to the first law of thermodynamics: $\Delta U = q + W$.$\Delta U = 800 \ J + (-405.3 \ J) = 394.7 \ J$.Rounding to the nearest provided option,the change in internal energy is $394.95 \ J$.

$A$ gas in a frictionless piston cylinder at a constant pressure of $1 \ atm$ expands from a volume of $2 \ L$ to $6 \ L$. 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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