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Q. A gas expands against a constant external pressure of $2.00$ atm, increasing its volume by $3.40 L.$ Simultaneously, the system absorbs $400 J$ of heat from its surroundings. What is $\Delta E$, in joules, for this gas?

Thermodynamics

Solution:

Given, $P = 2.00 \,atm,$
$\Delta V = 3 .4 \,L$
$q = 400\, J$
Work done by gas $w = -P\Delta V$
$= - 2 \times 3.4$
$= - 6.8 \,L \,atm$
$= - 6.8 \times 101.32\, J$
$= - 688.976 \,J$
$\therefore \Delta E=q+\left(-w\right)$
$=400+\left(-688.976\right)$
$=-288.976\approx-289 J$