One mole of an ideal gas at 300 K in thermal | vedclass

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(C) The change in entropy of the surroundings is given by $\Delta S_{	ext{surr}} = -\frac{q_{	ext{sys}}}{T}$.Since the gas expands against a constan

Vedclass · Accepted Answer

$-1.013$

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(C) The change in entropy of the surroundings is given by $\Delta S_{	ext{surr}} = -\frac{q_{	ext{sys}}}{T}$.Since the gas expands against a constant external pressure $P_{	ext{ext}}$,the work done by the gas is $W = -P_{	ext{ext}}(V_2 - V_1)$.According to the first law of thermodynamics,$\Delta U = q + W$. For an isothermal process of an ideal gas,$\Delta U = 0$,so $q = -W = P_{	ext{ext}}(V_2 - V_1)$.Given $P_{	ext{ext}} = 3.0 \ atm$,$V_1 = 1.0 \ L$,$V_2 = 2.0 \ L$,and $T = 300 \ K$.$q = 3.0 \ atm 	imes (2.0 \ L - 1.0 \ L) = 3.0 \ L \ atm$.Converting to Joules: $q = 3.0 	imes 101.3 \ J = 303.9 \ J$.The heat absorbed by the surroundings is $-q = -303.9 \ J$.Therefore,$\Delta S_{	ext{surr}} = \frac{-303.9 \ J}{300 \ K} = -1.013 \ J \ K^{-1}$.

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