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2 years ago

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An monatomic ideal gas at 300 K has a volume of 15 liters at a pressure of 15 atm. Calculate a. The final volume of the system b

. The work done by the system c. The heat entering or leaving the system d. The change in the internal energy e. The change in the enthalpy when the gas undergoes i. A reversible isothermal expansion to a pressure of 10 atm ii. A reversible adiabatic expansion to a pressure of 10 atm The constant-volume molar heat capacity of the gas, Cy, has the value 1.5 R.

1 answer:

Neporo4naja [7]2 years ago

3 0

Answer:

i) a) V2 = 22.5 L

b) W = 9249.451 J

c) Q = - 9249.451 J

d) ΔU = 0 J/mol

e) ΔH = 0 J/mol

ii) a) V2 = 19.154 L

b) W = 3390.346 J

c) Q = 0 J

d) ΔU = - 562.124 J/mol

e) ΔH = - 936.884 J/mol

Explanation:

an monoatomic ideal gas:

∴ PV = RTn

contants molar heat capacity:

∴ Cp,n = 2.5 R

∴ Cv,n = 1.5 R

i) a reversible isothermal expansion:

∴ T1 = T2 = 300 K

∴ P1 = 15 atm

∴ V1 = 15 L

∴ P2 = 10 atm

⇒ n = P1V1 / RT1 = ((15)*(15)) / ((0.082)*(300)) = 9.146 mol

a) V2 = RT2n/P2 = ((0.082)*(300)*(9.146)) / 10 = 22.5 L

b) W = nRT Ln (P1/P2)

⇒ W = (9.146)*(8.314)*(300) Ln(15/10)

⇒ W = 9249.451 J

c) Q = - W = - 9249.451 J

d) ΔU = CvΔT

∴ ΔT = 0

⇒ ΔU = 0

e) H = U + PV.....ideal gas

⇒ ΔH = ΔU + ΔPV = 0 + (P1V1 - P2V2) = 0 + 0 = 0

ii) a reversible adiabatic expansion:

∴ P2 = 10 atm

a) P1(V1)∧γ = P2(V2)∧γ....reversible adiabatic

∴ γ = Cp,n / Cv,n = 2.5R / 1.5R = 1.666

⇒ (V2)∧γ = ((15)(15)∧γ) / 10

⇒ (V2)∧γ = 136.849

⇒ V2 = ( 136.849 )∧(1/1.666)

⇒ V2 = 19.154 L

b) W = P1V1 - P2V2

⇒ W = ((15)*(15)) - ((10)*(19.154)) = 225 - 191.54 = 33.46 atm.L

⇒ W = 33.46 atm.L * (Pa/9.8692 E-6atm) * (m³/1000L) = 3390.346 Pa.m³

⇒ W = 3390.346 J

c) Q = 0 J...adiabatic

d) ΔU = Cv,n*ΔT

∴ T2/T1 = (V1/V2)∧(R/Cv,n)

⇒ T2 = T1 * [ (V1/V2)∧(R/Cv,n) ]

∴R/Cv,n = R/1.5R = 1/1.5= 0.666

⇒ T2 = 300K * [(15/19.154)∧(0.666)]

⇒ T2 = 254.925 K

⇒ ΔU = Cv,n*ΔT = 1.5*(8.314)*(254.925 - 300) = - 562.124 J/mol

e) ΔH = ∫Cp,n*δT = Cp,n*ΔT.....constant molar heat capacity

⇒ ΔH = 2.5*(8.314)*(254.925 - 300) = - 936.884 J/mol

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Answer:

1.053×10²⁴ atoms of gold

Explanation:

Hello,

Gold nugget are usually the natural occurring gold and they contain 85% - 90% weight of pure gold.

In this question, we're required to find the number of atoms in 344.75g of a gold nugget.

We can use mole concept relationship between Avogadro's number and molar mass.

1 mole = molar mass

Molar mass of gold = 197 g/mol

1 mole = Avogadro's number = 6.022 × 10²³ atoms

Number of mole = mass / molar mass

Mass = number of mole × molar mass

Mass = 1 × 197

Mass = 197g

197g is present in 6.022×10²³ atoms

344.75g will contain x atoms

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2 years ago

A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)

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Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

$\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }$

<u>Given:</u>

A mysterious white powder could be,

powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

The solute = the powder
The solvent = ethanol
The freezing point of the solvent = −114.6°C
The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

Mass of solute = 82 mg = 0.082 g
Mass of solvent = density x volume, i.e., $\boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg \ }$

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

$\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

Now we combine it with the formula of freezing point depression.

$\boxed{ \ \Delta T_f = K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

$\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }$

$\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }$

$\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }$

We get $\boxed{ \ Mr = 153.65 \ }$

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>

The molality and mole fraction of water brainly.com/question/10861444
About the mass and density of ethylene glycol as an antifreeze brainly.com/question/4053884
About the solution as a homogeneous mixture brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

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2 years ago

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You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain

IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of $O_2$ :-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{49.8\ g}{32\ g/mol}$

$Moles_{O_2}= 1.55625\ mol$

Since pressure and volume are constant, we can use the Avogadro's law as:-

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

Given ,

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

$\frac {V_1}{n_1}=\frac {V_2}{n_2}$

$\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}$

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of $CO_2$ = 44.0 g/mol

Mass of $CO_2$ = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>

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Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t

Dafna11 [192]

Answer:

The plane with aluminium can lift more mass of passangers than the plane of steel.

Explanation:

The total mass the airplane canc lift is:

$m_{tot}=m_{fuselage}+m_{passangers}$

For aluminium:

$m_{tot}=m_{fus-Al}+m_{pas-Al}$

$m_{fus-Al}=\delta _{Al}*V_{fuselage}$

and

$V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]$

where:

L is lenght
D is diameter
e is thickness

$m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}$

For steel (same procedure):

$m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel$

Knowing that the total mass the airplane can lift is constant and that aluminum has a lower density than the steel, we can afirm that the plane with aluminium can lift more mass of passangers.

Also you can estimate an average weight of passanger to estimate a number of passangers it can lift.

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2 years ago

Suppose there is 0.63 g of HNO3 per 100 mL of a particular solution. What is the concentration of the HNO3 solution in moles per

Vadim26 [7]

Answer:

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