Answer
(a) $ n = 21.5 mol$
(b) $\Delta U = -1.79 \times 10^4 J $
(c) $W = -7100 J $
(d) $Q = -1.79 \times 10^4 J $
Work Step by Step
(a) Number of moles of the gas
$n = \frac{Q}{C_p \Delta T }$ and $C_P = 29.07 J/mol ⋅K $ for nitrogen gas.
$n = \frac{-2.5 \times 10^4 J}{(29.07 J/mol ⋅K
)(-40K)} $
$ n = 21.5 mol$
(b) Change in the internal energy of the gas,
$\Delta U = nC_VΔT$
$\Delta U = (21.5 mol)(20.76 J/mol.K)(-40)$
$\Delta U = -1.79 \times 10^4 J $
(c) The work done is given by The first law of thermodynamics.
$W= Q - \Delta U $
$W= -2.5 \times 10^4 J -(-1.79 \times 10^4 J) $
$W = -7100 J $
(d) If the volume was constant, $\Delta V = 0$ then,
$ W = p\Delta V $
$ W = p(0) $
$ W = 0 $
So the heat is
$Q = \Delta U + W$
$Q = -1.79 \times 10^4 J + 0 $
$Q = -1.79 \times 10^4 J $
The heat is applied to the system, indicated by the negative sign.
