Question

A gas is allowed to expand at constant temperature from a volume of $1.0L$ to $10.1L$ against an external pressure of $0.50atm$ . If the gas absorbs $250J$ of heat from the surroundings, what are the values of $q,w$ and $\Delta E$ ? (Given $1Latm=101.0J$ )

• A.

  q	w	$\Delta E$
  250 J	- 460 J	- 210 J

• B.

  q	w	$\Delta E$
  -250 J	- 460 J	- 710 J

• C.

  q	w	$\Delta E$
  250 J	460 J	710 J

• D.

  q	w	$\Delta E$
  -250 J	460 J	210 J

Answer 1

Answer: (A)

Since heat is absorbed, $q=+250\text{ J}$
Work done $=-P\left(V_{2} - V_{1}\right)$
$=-50\text{ atm}\times \left(\text{10.1 L} - 1.0 \text{ L}\right)$
$=-0.50\text{ atm}\times 9.1\text{ L}$
$=-0.50\times 9.1\text{ L atm}\times \frac{101.0 \text{ J}}{1 \text{ L atm}}=-459.5\text{ 5J}$ $\sim eq$ $-460.0$
Also $\Delta E=q+w=250\text{ J}-460\text{ J}=-210\text{ J}$