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## Suppose that 4.4 moles of a monatomic ideal gas (atomic mass = 7.9 × 10-27 kg) are heated from 300 K to 500 K at a constant volume of 0.44 m

Question

Suppose that 4.4 moles of a monatomic ideal gas (atomic mass = 7.9 × 10-27 kg) are heated from 300 K to 500 K at a constant volume of 0.44 m3. It may help you to recall that CV = 12.47 J/K/mole and CP = 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.1) How much energy is transferred by heating during this process?

2) How much work is done by the gas during this process?

3) What is the pressure of the gas once the final temperature has been reached?

4) What is the average speed of a gas molecule after the final temperature has been reached?

5) The same gas is now returned to its original temperature using a process that maintains a constant pressure. How much energy is transferred by heating during the constant-pressure process?

6) How much work was done on or by the gas during the constant-pressure process?

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Physics
3 years
2021-09-02T19:36:31+00:00
2021-09-02T19:36:31+00:00 1 Answers
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## Answers ( )

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 ΔTwhere,

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)ΔQ₁ = 10.97 x 10³ J = 10.97 KJ2)Since, work done by gas is given as:

W₁ = PΔVwhere,

ΔV = 0, due to constant volume

Therefore,

W₁ = 0 J4)The average kinetic energy of a gas molecule is given as:

K.E = (3/2)KTbut, K.E is also given by:

K.E = (1/2)mv²Comparing both equations:

(1/2)mv² = (3/2)KTmv² = 3KTv = √(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)]v = 1618.72 m/s3)From kinetic molecular theory, we know that or an ideal gas:

P = (1/3)ρv²where,

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)²P = 41.66 x 10³ Pa = 41.66 KPa5)The heat transfer for a constant pressure process is given by the formula:

ΔQ₂ = n Cp ΔTwhere,

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)ΔQ₂ = – 18.29 x 10³ J = – 18.29 KJNegative sign shows heat flows from system to surrounding.6)From Charles’ Law, we know that:

V₁/T₁ = V₂/T₂V₂ = (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 m³Therefore,

ΔV = V₂ – V₁ = 0.264 m³ – 0.44 m³ = – 0.176 m³Hence, the work done , will be:

W₂ = PΔV = (41.66 KPa)(- 0.176 m³)W₂ = – 7.33 KJNegative sign shows that the work is done by the gas