Answer:
1) ΔQ₁ = 10.97 x 10³ J = 10.97 KJ
2) W₁ = 0 J
3) P = 41.66 x 10³ Pa = 41.66 KPa
4) v = 1618.72 m/s
5) ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ
6) W₂ = - 7.33 KJ
Explanation:
1)
The heat transfer for a constant volume process is given by the formula:
ΔQ₁ = ΔU = n Cv ΔT
where,
ΔQ₁ = Heat transfer during constant volume process
ΔU = Change in internal energy of gas
n = No. of moles = 4.4 mol
Cv = Molar Specific Heat at Constant Volume = 12.47 J/mol.k
ΔT = Change in Temperature = T₂ - T₁ = 500 k - 300 k = 200 k
Therefore,
ΔQ₁ = (4.4 mol)(12.47 J/mol.k)(200 k)
<u>ΔQ₁ = 10.97 x 10³ J = 10.97 KJ</u>
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2)
Since, work done by gas is given as:
W₁ = PΔV
where,
ΔV = 0, due to constant volume
Therefore,
<u>W₁ = 0 J</u>
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4)
The average kinetic energy of a gas molecule is given as:
K.E = (3/2)KT
but, K.E is also given by:
K.E = (1/2)mv²
Comparing both equations:
(1/2)mv² = (3/2)KT
mv² = 3KT
v = √(3KT/m)
where,
v = average speed of gas molecue = ?
K = Boltzman Constant = 1.38 x 10⁻²³ J/k
T = Absolute Temperature = 500 K
m = mass of a molecule = 7.9 x 10⁻²⁷ kg
Therefore,
v = √[(3)(1.38 x 10⁻²³ J/k)(500 k)/(7.9 x 10⁻²⁷ kg)]
<u>v = 1618.72 m/s</u>
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3)
From kinetic molecular theory, we know that or an ideal gas:
P = (1/3)ρv²
where,
P = pressure of gas = ?
m = Mass of Gas = (Atomic Mass)(No. of Atoms)
m = (Atomic Mass)(Avogadro's Number)(No. of Moles)
m = (7.9 x 10⁻²⁷ kg/atom)(6.022 x 10²³ atoms/mol)(4.4 mol)
m = 0.021 kg
ρ = density = mass/volume = 0.021 kg/0.44 m³ = 0.0477 kg/m³
Therefore,
P = (1/3)(0.0477 kg/m³)(1618.72 m/s)²
<u>P = 41.66 x 10³ Pa = 41.66 KPa</u>
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5)
The heat transfer for a constant pressure process is given by the formula:
ΔQ₂ = n Cp ΔT
where,
ΔQ₂ = Heat transfer during constant pressure process
n = No. of moles = 4.4 mol
Cp = Molar Specific Heat at Constant Pressure = 20.79 J/mol.k
ΔT = Change in Temperature = T₂ - T₁ = 300 k - 500 k = -200 k
Therefore,
ΔQ₂ = (4.4 mol)(20.79 J/mol.k)(-200 k)
<u>ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ</u>
<u>Negative sign shows heat flows from system to surrounding.</u>
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6)
From Charles' Law, we know that:
V₁/T₁ = V₂/T₂
V₂ = (V₁)(T₂)/(T₁)
where,
V₁ = 0.44 m³
V₂ = ?
T₁ = 500 K
T₂ = 300 k
Therefore,
V₂ = (0.44 m³)(300 k)/(500 k)
V₂ = 0.264 m³
Therefore,
ΔV = V₂ - V₁ = 0.264 m³ - 0.44 m³ = - 0.176 m³
Hence, the work done , will be:
W₂ = PΔV = (41.66 KPa)(- 0.176 m³)
<u>W₂ = - 7.33 KJ</u>
<u>Negative sign shows that the work is done by the gas</u>
(1)
is the spring constant and 
is the compression of the spring. If we substitute the numbers, we find:
