One mole of an ideal gas at 350 , K is in a 2 | vedclass

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(C) For an isothermal process,the heat exchanged with the surroundings is $q = -w = P_{ext} \Delta V$. Given $P_{ext} = 4 \, atm$ and $\Delta V = (3.0

Vedclass · Accepted Answer

$-1$

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(C) For an isothermal process,the heat exchanged with the surroundings is $q = -w = P_{ext} \Delta V$. Given $P_{ext} = 4 \, atm$ and $\Delta V = (3.0 \, L - 2.0 \, L) = 1.0 \, L$. $q = 4 \, atm 	imes 1.0 \, L = 4 \, L \, atm$. Using the conversion $1 \, L \, atm = 101.3 \, J$,we get $q = 4 	imes 101.3 = 405.2 \, J$. The entropy change of the surroundings is given by $\Delta S_{surr} = -\frac{q_{sys}}{T}$. Since the process is isothermal and the system absorbs heat from the surroundings,$q_{surr} = -q_{sys} = -405.2 \, J$. $\Delta S_{surr} = \frac{q_{surr}}{T} = \frac{-405.2 \, J}{350 \, K} \approx -1.158 \, J \, K^{-1}$. The nearest integer is $-1$.

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