274 mL TNT C7H5N3O6 2.00 mol 1.0 atm 448 L gas PV work kJ

274 mL TNT C7H5N3O6 2.00 mol 1.0 atm 448 L gas PV work kJ

 

 

The explosion of 2.00 mol of solid trinitrotoluene (TNT; C7H5N3O6)

with a volume of approximately 274 mL produces gases

with a volume of 448 L at room temperature and 1.0 atm pressure.

How much PV work in kilojoules is done during the explosion?

2C7H5N3O6(s) → 12CO(g) + 5H2(g) + 3N2(g) + 2C(s)

 

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w = –PdV

( 참고 https://ywpop.tistory.com/5281 )

 

= –(1.0 atm) (448 – 0.274 L)

= –447.726 atm•L

 

 

 

기체상수, R = 8.314 J/mol•K = 0.08206 atm•L/mol•K

---> 8.314 J = 0.08206 atm•L

 

 

 

(–447.726 atm•L) (8.314 J / 0.08206 atm•L)

= –45362 J

 

 

 

답: –45.36 kJ

 

 

 

 

 

 

 

 

[키워드] TNT PV work 기준문서