Answer
(a)$T_1 =325 K$
$T_2 =456 K$
(b) $W =3.29 \times 10^3J $
(c) $Q =8.17 \times 10^3 J$
(d) $\Delta U = 4.88 \times 10^3 J $
Work Step by Step
(a) The temperature at initial point
$T_1 = \frac{pV_1}{nR}$
$T_1 = \frac{(2.50 atm)(1.013 \times 10^5 Pa/atm)(3.20 \times 10^{-2} m^3)}{(3.0)(8.314 J/mol.K)}$
$T_1 =325 K$
The temperature at final point
$T_2 = \frac{(2.50 atm)(1.013 \times 10^5 Pa/atm)(4.50 \times 10^{-2} m^3)}{(3.0)(8.314 J/mol.K)}$
$T_2 =456 K$
(b) $W = p\Delta V$
$W = (2.50 atm)(1.013 \times 10^5 Pa/atm) (4.50 \times 10^{-2} m^3 - 3.20 \times 10^{-2} m^3) $
$W = 3292.25 J = 3.29 \times 10^3J $
(c) $Q = nC_p \Delta T$
$Q = (3.0) (\frac{5}{2} R) \Delta T$
$Q =(3.0) (\frac{5}{2} ) (8.314 J/mol.K) (456 K - 325 K) $
$Q =8.17 \times 10^3 J$
(d) $\Delta U = Q - W$
$\Delta U = 8.17 \times 10^3 J - 3.29 \times 10^3 J $
$\Delta U = 4.88 \times 10^3 J $
