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Arada [10]
3 years ago
15

A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas (both gases are monatomic) at 27

°C and 200 kPa. Now the gas expands at constant pressure until its volume doubles. The amount of heat transfer to the gas mixture is, in MJ (round to nearest integer; for example if the answer is 53.7MJ, write 54; if the answer is 52.1MJ write 52; do not include the units in your answer)
Engineering
2 answers:
lidiya [134]3 years ago
5 0

Answer:

Q = 62    ( since we are instructed not to include the units in the answer)

Explanation:

Given that:

n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol

T_1 = 27^0 \ C = ( 27+273)K =  300 K

P_1 = 200 \ kPa

Q = ???

Now the gas expands at constant pressure until its volume doubles

i.e if V_1 = x\\V_2 = 2V_1

Using Charles Law; since pressure is constant

V \alpha T

\frac{V_2}{V_1}  =\frac{T_2}{T_1}

\frac{2V_1}{V_1}  =\frac{T_2}{300}

T_2 = 300*2\\T_2 = 600

mass of He =number of moles of He × molecular weight of He

mass of He = 3 kg  × 4

mass of He = 12 kg

mass of Ar =number of moles of Ar × molecular weight of Ar

mass of He = 7 kg  × 40

mass of He = 280 kg

Now; the amount of  Heat  Q transferred = m_{He}Cp_{He} \delta T  + m_{Ar}Cp_{Ar} \delta T

From gas table

Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar}  = 0.5203 \  kJ/Kg/K

∴ Q = 12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)

Q = 62.389 *10^6

Q = 62 MJ

Q = 62    ( since we are instructed not to include the units in the answer)

vfiekz [6]3 years ago
4 0

Answer:

Q = 62

Explanation:

The solution is obtained by manipulating the energy balance of the system:

Q = W + ΔU

⇒ Q = P*(V₂ - V₁) + mcv*(T₂ - T₁)

⇒ Q = P*V₁ + m*(((mcv)He /m) + ((mcv)Ar /m)*((P*V₂/(mR) - T₁)

⇒ Q = mRT₁ + ((mcv)He + (mcv)Ar)*(2P*V₁/(mR) - T₁)

⇒ Q = T₁*(Nm*Ru + (mcv)He + (mcv)Ar)

⇒ Q = T₁*(Nm*Ru + (MNcv)He + (MNcv)Ar)

where

T₁ = (27 + 273) K = 300 K

Nm = (7 + 3) kmol = 10 kmol = 10⁴ mol

Ru = 8.314 J*K⁻¹*mol⁻¹

(MNcv)He = (4 g*mol⁻¹)*(3*10³ mol)*(3.1156 kJ*Kg⁻¹*K⁻¹)*(1 kg/10³ g)*(1 MJ/10³ kJ) = 0.0373872 MJ*K⁻¹

(MNcv)Ar = (40 g*mol⁻¹)*(7*10³ mol)*(0.3122 kJ*Kg⁻¹*K⁻¹)*(1 kg/10³ g)*(1 MJ/10³ kJ) = 0.087416 MJ*K⁻¹

Finally, we get

⇒ Q = 300*(10*8.314 + 4*3*3.1156 + 40*7*0.3122)*10⁻³

⇒ Q = 62

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What are the transfer of energy for the 6 simple machine? Ex: inclined plane is gravitational potential energy to mechanical ene
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Answer:

I. Inclined plane: GPE to KE.

II. Screw: KE to GPE.

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IV. Lever: GPE to KE.

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VI. Pulley: GPE to KE.

Explanation:

A simple machine can be defined as a type of machine with no moving parts but can be used to perform work.

Basically, a simple machine allows for the transformation of energy into work. The six simple machines are;

I. <u>Inclined plane</u>: gravitational potential energy to mechanical energy (kinetic energy). It is set at an angle and then used to lift an object.

II. <u>Screw</u>: kinetic energy to potential energy. It is held in position and then used to tighten two or more objects together.

III. <u>Wheel and axle</u>: gravitational potential energy to mechanical energy (kinetic energy). It comprises of two circular objects which aids in the motion of an object such as a car.

IV. <u>Lever</u>: gravitational potential energy to mechanical energy (kinetic energy). It comprises of a pivoted bar at a fixed point (fulcrum) that allows it to be used for lifting heavy objects.

V. <u>Wedge</u>: gravitational potential energy to mechanical energy (kinetic energy). It can be used to split objects into halves.

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Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

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8 0
2 years ago
An excited electron in an Na atom emits radiation at a wavelength 589 nm and returns to the ground state. If the mean time for t
11Alexandr11 [23.1K]

Answer:   Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm

                length of the photon emitted: 6.0 m

Explanation:

The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.

Recall the Heisenberg's uncertainty principle:-

                                 ∆t∆E ≈ h ( Planck's Constant)

The transition time ∆t corresponds to the energy that is ∆E

E=h/t = \frac{(1/2\pi)*6.626*10x^{-34} J.s}{20*10x^{-9} } = 5.273*10x^{-27} J =  3.29* 10^{-8} eV.

The corresponding uncertainty in the emitted frequency ∆v is:

∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)=  7.958 × 10^6 s^-1

To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate

                                                    λ = c/v

                                                    dλ/dv = -c/v² = -λ²/c

Therefore,

      ∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)

                                 =  9.20 × 10⁻¹⁵ m or 9.20 fm

     The length of the photon (<em>l)</em> is

l = (light velocity) × (emission duration)

  = (3.0 × 10⁸  m/s)(20 × 10⁻⁹ s) = 6.0 m          

                                                   

6 0
3 years ago
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