A gas expands from a volume of 1 text{ dm}^3 | vedclass

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(A) Step $1$: Calculate the work done by the system using $w = -P_{ext} \Delta V$.Step $2$: Convert units: $1 	ext{ bar} = 10^5 	ext{ Pa}$ and $1

Vedclass · Accepted Answer

$125$

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(A) Step $1$: Calculate the work done by the system using $w = -P_{ext} \Delta V$.Step $2$: Convert units: $1 	ext{ bar} = 10^5 	ext{ Pa}$ and $1 	ext{ dm}^3 = 10^{-3} 	ext{ m}^3$.Step $3$: $\Delta V = 1.25 	ext{ dm}^3 - 1 	ext{ dm}^3 = 0.25 	ext{ dm}^3 = 0.25 	imes 10^{-3} 	ext{ m}^3$.Step $4$: $w = -(10^5 	ext{ Pa}) 	imes (0.25 	imes 10^{-3} 	ext{ m}^3) = -25 	ext{ J}$.Step $5$: Apply the first law of thermodynamics: $\Delta U = q + w$.Step $6$: Given $q = +150 	ext{ J}$ (heat supplied to the system).Step $7$: $\Delta U = 150 	ext{ J} - 25 	ext{ J} = 125 	ext{ J}$.

$A$ gas expands from a volume of $1 \text{ dm}^3$ to $1.25 \text{ dm}^3$ against a constant external pressure of $1 \text{ bar}$. Calculate the change in internal energy $(\Delta U)$ if $150 \text{ J}$ of heat is supplied to the system. (in $\text{ J}$)

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