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Volgvan
3 years ago
8

A helium gas balloon is expanded to 78.0 L, while the pressure is held constant at 0.37 atm. If the work done on the gas mixture

was 28.2 J, what was the initial volume (in liters) of the helium gas balloon. (1 L. atm?
Chemistry
2 answers:
Elenna [48]3 years ago
8 0

Answer:

77.248 L

Explanation:

From the question,

Work done on the gas mixture is given as,

W = PΔV.................. Equation 1

Where W = work done, P = pressure of the the gas, ΔV = Change in volume of the gas.

make ΔV the subject of the equation

ΔV = W/P..................... Equation 2

Given: W = 28.2 J, P = 0.37 atm = (0.37×101325) N/m² = 37490.25 N/m²

Substitute into equation 2

ΔV = 28.2/37490.25

ΔV = 0.000752 m³

ΔV = 0.752 L

But,

ΔV = V₂-V₁................. Equation 3

Where V₂ = Final volume of the helium gas, V₁ = Initial volume of the helium gas

make V₁ the subject of the equation

V₁ = V₂-ΔV................ Equation 4

Given: V₂ = 78 L.

Substitute into equation 4

V₁ = 78-0.752

V₁ = 77.248 L

vlada-n [284]3 years ago
5 0

Answer:

The initial volume of the helium gas balloon was 1.78 L

Explanation:

Step 1: Data given

Volume of the balloon is expanded to 78.0 L

The pressure is held constant at 0.37 atm

If the work done on the gas mixture was 28.2 J

Step 2: Calculate the initial volume

W = pΔV

⇒W = the work done on the gas = 28.2 J

⇒p = the pressure = 0.37 atm

⇒ΔV = the change in volume = V2 - V1 = 78.0 L - V1

W = 0.37 * ( 78.0 - V1)

28.2 J = 0.37 * ( 78.0 - V1)

28.2 J = 28.86 - 0.37V1

-0.66 = -0.37V1

V1 = 1.78 L

The initial volume of the helium gas balloon was 1.78 L

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Answer:

67.824

Explanation:  You want to use the combined gas law equation (P1*V1)/(n1*T1)=(P2*V2)/(n2*T2). So first cross out what remains constant, so volume(V) and I assume moles (since it was not mentioned as a change). Then you can solve algebraically for the answer!

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Neon is compressed from 100 kPa and 16°C to 500 kPa in an isothermal compressor. Determine the change in the specific volume and
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Answer:

\Delta v=-0.952m^3/kg

\Delta h=0

Explanation:

In this case, since neon could be considered as an ideal gas, the specific volumes at the first and second state are respectively:

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3 0
4 years ago
How much energy is required to heat 87.1 g acetone (molar mass=58.08 g/mol) from a solid at -154.0°C to a liquid at -42.0°C? The
WARRIOR [948]

Answer:

The answer to the question above is

The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C = 29.36 kJ

Explanation:

The given variables are

ΔHfus = 7.27 kJ/mol

Cliq = 2.16 J/g°C

Cgas = 1.29 J/g°C

Csol = 1.65 J/g°C

Tmelting = -95.0°C.

Initial temperature = -154.0°C

Final temperature = -42.0°C?

Mass of acetone = 87.1 g

Molar mass of acetone = 58.08 g/mol

Solution

Heat required to raise the temperature of solid acetone from -154 °C to -95 °C or 59 °C is given by

H = mCsolT = 87.1 g* 1.65 J/g°C* 59 °C = 8479.185 J

Heat required to melt the acetone at -95 °C = ΔHfus*number of moles =

But number of moles = mass÷(molar mass) = 87.1÷58.08 = 1.5

Heat required to melt the acetone at -95 °C =1.5 moles*7.27 kJ/mol = 10.905 kJ

The heat required to raise the temperature to -42 degrees is

H = m*Cliq*T = 87.1 g* 2.16 J/g°C * 53 °C = 9971.21 J

Total heat = 9971.21 J + 10.905 kJ + 8479.185 J = 29355.393 J = 29.36 kJ

The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C is 29.36 kJ

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andrey2020 [161]
<span>Hi, friend.
Steepest - Being steep to the greatest degree.
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