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for the real gases in reaction 2Co(g) +O2(g)-2CO2(g) ∆H=-560 kJ.In 10 L rigid vessel at 500K the initial pressure is 70 bar and after the reaction it becomes 50 barr. The change in internal energy is

for the real gases in reaction 2Co(g) +O2(g)-2CO2(g) ∆H=-560 kJ.In 10 L rigid vessel at 500K the initial pressure is 70 bar and after the reaction it becomes 50 barr. The change in internal energy is

Grade:12th pass

2 Answers

Arun
25763 Points
2 years ago

We have,             ∆H          =             ∆U          +             ∆(PV)

=             ∆U          +             P2V2 – P1V1 —           —           (1)

As gases deviates from ideal gas behaviour, we cannot use,  ∆H                =             ∆U          +             ∆nRT

Thus                                      ∆(PV)    =             P2V2 – P1V1

=             40×1 – 70×1

=             – 30 L-atm

=             – 3.0 kJ

Therefore, by (1)             – 560     =             ∆U – 3.0

Therefore,                          ∆U          =             – 557 kJ

technical gaming artist
24 Points
one year ago
At constant volume ∆H = ∆U + V∆P
 
=> -560 = ∆U + 10 x (-30) x 0.1
=>∆U=-530 kJ
Multipled the value with 0.1 to get answer in Joules

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