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(C) For an isothermal process of an ideal gas, the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics, $\Delta U =

Vedclass · Accepted Answer

$6$

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(C) For an isothermal process of an ideal gas, the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics, $\Delta U = q + w$, so $q = -w$.The work done on the system during compression against a constant external pressure is given by $w = -P_{	ext{ext}} \Delta V$.Here, $P_{	ext{ext}} = 4 	ext{ bar} = 4 	imes 10^5 	ext{ Pa}$, $V_1 = 25 	ext{ dm}^3 = 25 	imes 10^{-3} 	ext{ m}^3$, and $V_2 = 10 	ext{ dm}^3 = 10 	imes 10^{-3} 	ext{ m}^3$.$\Delta V = V_2 - V_1 = (10 - 25) 	imes 10^{-3} 	ext{ m}^3 = -15 	imes 10^{-3} 	ext{ m}^3$.$w = -(4 	imes 10^5 	ext{ Pa}) 	imes (-15 	imes 10^{-3} 	ext{ m}^3) = 6000 	ext{ J} = 6 	ext{ kJ}$.Since $q = -w$, the heat released is $q = -6 	ext{ kJ}$.The quantity of heat released is $6 	ext{ kJ}$.

Calculate the quantity of heat released from a system when $2 \text{ moles}$ of an ideal gas are compressed isothermally from a volume of $25 \text{ dm}^3$ to $10 \text{ dm}^3$ at a constant external pressure of $4 \text{ bar}$. (in $\text{ kJ}$)

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