For state 1

PV = nRT

1 * 10 = n(0.0821)(300)

n = 0.4

For state 2

PV = nRT

\( 4 * 5 = 0.4 * (0.0821)(T_{2}) \)

T2 = 600 K

\( \Delta U = q + W \)

= \( 50 (\Delta T) + (-P_{ex})(\Delta V) \)

= \( 50 (600 – 300)J – (1atm) (3L) \)

= 15000 – 300

= 14700

\( \Delta U = 14700 J \)

\( \Delta H = \Delta U + \Delta (PV) \)

= \( 14700 + (5 * 4 – 10)atm L \)

= 15700

\( \Delta H = 15.7 kJ \)

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