Two moles of molecular hydrogen (H2) react with 1 mole of molecular oxygen (O2)

Question

Two moles of molecular hydrogen (H2) react with 1 mole of molecular oxygen (O2) to produce 2 moles of water (H2O) together with an energy release of 241.8 kJ/mole of water. Suppose a spherical vessel of radius 0.500 m contains 14.4 moles of H2 and 7.2 moles of O2 at 20.0 *C. a) What is the initial pressure in the vessel? b) What is the initial internal energy of the gas? c) Suppose a spark ignites the mixture and the gases burn completely into water vapor. How much energy is produced? d) Find the temperature and pressure of the steam, assuming it's an ideal gas. e) Find the mass of steam and then calculate the steam's density. f) If a small hole were put in the sphere, what would be the initial exhaust velocity of the exhausted steam if spewed out into a vaccuum (use Bernoulli's equation).

what is the formula for the initial internal energy and how do we found it here

Explanation / Answer

According to the question

V=4Pi*R^3/3=0.52(m3).
We know that:
PV=nRT.
so P=nRT/V=(14.4+7.2)*8.31*293/0.52=1e5)

Then,
B). initial internal energy.
(14.4+7.2)*(5R/2)*293=1.31e5(J)

Then,
C) released energy by the combination.
241.8e3*14.4=34.8e5(J).
D) Then,
mass of water.
14.4*18e-3=0.26(kg).
That's
34.8e5=0.26*4200*(100-20)+0.26*2260e3+0.26*2080*(T-100)
So

T=5286(C)=5560(K).

PV=nRT so P=nRT/V=12.8e5(Pa)

Then,
E) mass of steam = mass of water.
M=0.26kg.
steam density M/V=0.26/0.52=0.5(kg/m3).
F) Then,
P=D*v^2/2
so v=sqrt(2P/D)=2262(m/s)

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