a cylinder containing hydrogen gas physics question?

Question

A cylinder contains 270 L of hydrogen gas (H2 at 0.0° C) and a pressure of 10 atm. How much energy is required to raise the temperature of this gas to 25°C?

Answer 1

Hydrogen H2 consists of diatomic molucules, which molecules interact only by collisons. As such, hydrogen is very good ideal gas obeying the law
PV = nRT

Any ideal gas has zero potential energy, all its energy is kinetic.

Translational kinetic energy of any ideal gas is
K = 3/2 RT
(This is exact formula, which can be easily derived from virial of forces acting on the the gas:
KE = - 1/2 virial =
= 1/2 x 4πRadius² x Radius x Pressure =
= 3/2 (4/3 πRadius³) x Pressure =
= 3/2 Volume x Pressure =
= 3/2 RT
)

So far so good. Your textbook states that in addition to translational kinetic energy there is also rotational kinetic energy.
Monoatomic gases cannot rotate 'because they are material points'.
Diatomic gases can rotate about two short axes, but not about the long axis, 'because they are material points'.

This explantion is more or less reasonable, as long as you do not touch diatomic hydrogen, which hydrogen spits on everything firmly established (and also rewards close attention with many Nobel prizes awarded).

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Diatomic gases can rotate about shorter axes, becuase the quantum of angular momentum is , and room temperatures exceed minimal energy of rotation
Emin = 1/2 ² / Ixx = ²/ md² >> kB T

For all gases but hydrogen. Note 'm' in the denominator, the mass of hydrogen is the smallest m(H) = 1 atomic unit, and minimal roatational energy of hydrogen molecule is the largest. In fact this minimal energy corresponds to about 100K temperature.

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Now it is time to do your homework.
Since you do not state explicitly, I will assume that gas is under a piston and under constant pressure.
Equation of state of ideal gas:
PV = nRT

Specific heat of diatomic ideal gas (except hydrogen, but lets ignore that):
Cv = 3/2R + 2/2R = 5/2 R

Energy of diatomic ideal gas:
E = nCvT = 5/2 RT = 5/2 PV

Enthalpy of diatomic ideal gas:
H = E + PV = 7/2 PV

Energy required to raise the temperature of this gas by ΔT = +25°C = +25K:
Work = ΔE + PΔV = Δ(E + PV) = ΔH
Work = 7/2 PΔV = 7/2 nRΔT
Because PVo = nRTo we have
n = PVo/RTo, and
Work = 7/2 nRΔT = 7/2 PVo/RTo RΔT
Work = 7/2 PVo ΔT/To


Answer:
Work = 7/2 PVo ΔT/To =
http://en.wikipedia.org/wiki/Pressure
7/2 (1013250 Pa) 0.270m³ 25K / 273.15K =
87637 Joule
Note that gas constant R is not present in the answer.

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Now we first need distance d between the protons in order to calculate moment of inertia Ixx = 1/2 md². Lets use Bohr's approach, namely place one proton at point (0,0,-d/2), the second at point (0,0,+d/2) and make two electrons rotate in xy plane in orbits of radius r and opposite to each other. Angular momentum of each electron must be on authority of Bohr L = Mvr = . Balancing the forces:
PROTONS:
e²(1/d² - d/√(r² + (d/2)²)³) = 0
r = √3 d/2

ELECTRONS:
Mvr =
Mv²/r = ke²(3√3/4 1/r² - 1/4 1/r²)

M²v²r² = ²
Mv²r = ke²(3√3 - 1)/4

r = ²/(Mke²) 4/(3√3) = 4/(3√3 - 1) Rbohr,
where
Rbohr = ²/(Mke²) = 0.529177 A

Finally the distance d between the protons is
d = r 2/√3 = 8/(9-√3) Rbohr = 0.58 Angstroem

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Wave function of a pair of protons is Ψ(r1,r2,s1,s2). Rewriting coordinates in terms of ceter of mass R=(r1+r2)/2 and relative position of
first proton to the center of mass r =(r1-r2)/2 we can separate translational motion and rotational motion:
Ψ = Ψt(R) Ψr(r,s1,s2)
Eigen states of rotataional motion have angular momenta
L = 0, 1, 2 ..., and their energies are
E(L) = (Lhat)²/2Ixx = L(L+1)²/(md²)
Lets denote
Eo = ²/(md²)
Eo = m_e/m_p (9-√3)]²/32 Erydberg
Eo = 0.0122 eV = 145 Kelvin

Orbital states with even angular momentum are symmeteric Ψr(r,s1,s2) = Ψr(-r,s1,s2), and therefore require anti-parallel spins of protons
S = S1 + S2 = 0 (this phase is called parahydrogen).

Orbital states with odd angular momentum are anti-symmeteric Ψr(r,s1,s2) = -Ψr(-r,s1,s2), and therefore require parallel spins of protons
S = S1 + S2 = 1 ( this phase is called orthohydrogen). Such spinior states are trice-degenrate: Sz = -1, 0, +1.

Therefore we have the following rotational eigen states:
L = 0, E = 0, 1 para-state Lz = 0
L = 1, E = 2Eo, 9 ortho-states (3x(Lz=-1,0,+1))
L = 2, E = 6Eo, 5 para-states (1x(Lz=-2,-1,0,+1,+2))
L = 3, E = 12Eo, 21 ortho-states
L = 4, E = 20Eo, 9 para-states
etc.

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Now we are ready to calculate internal energy corrsponding to rotation of hydrogen molecules.
Probability to find molecule in particular state of energy E is
p(E) = Po exp(-E/T)
1 = Σ p_i = Σ Po exp(-Ei/T)
Po = 1/Σ exp(-Ei/T)

Rotattional energy:
R(T) = Σ p_i Ei = Σ Ei exp(-Ei/T) / Σ exp(-Ei/T)

Thus we need to calculate sums
1/Po = Σ exp(-Ei/T) =
Σ (2n+1)exp(-2n(2n+1) Eo/T) +
+ 3 Σ(2n+3)exp(-(2n+1)(2n+2) Eo/T)

and
1/Eo Σ Ei exp(-Ei/T) =
Σ (2n+1) 2n(2n+1) exp(-2n(2n+1) Eo/T) +
+ 3 Σ(2n+3) (2n+1)(2n+2) exp(-(2n+1)(2n+2) Eo/T)


(to be continued)

Answer 2

To be able to keep the same pressure and temperature in both cylinders there must be a different amount of Hydrogen atoms in the first container then there are Oxygen atoms in the second. So in one container there will be less atoms than in the other meaning atoms in this container can move around quicker as they will not collide with other atoms as often as they do in the other container. But with this there are a couple of other factors that can affect an atoms speed: -The difference in size between the Hydrogen and Oxygen atoms -Different elements move at different speeds even if they are under the same temperature Hope this helps :)