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Oksi-84 [34.3K]
3 years ago
12

Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How m

uch heat is absorbed and how much work is done in the expansion ?
Chemistry
1 answer:
nlexa [21]3 years ago
4 0

Answer:

The work done and heat absorbed are both -8,1 kJ

Explanation:

The work done in an isobaric process is defined as:

W = -P (Vf - Vi)

Where P is pressure ( 10 atm)

Vf = 10 L

Vi = 2 L

Thus, <em>W = -80 atm×L ≡ -8,1 kJ</em>

This is the work done in expansion of the gas.  As the gas remains at the same temperature, there is no change in internal energy doing that all work was absorbed as heat.

I hope it helps!

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Answer: Elections

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Which of the following would increase friction?
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B. Sand.

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What represents the hemisphere
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3 years ago
Combustion of hydrocarbons such as nonane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosp
maksim [4K]

Answer:

Part 1: C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)

Part 2: Volume of CO₂ produced = 1223.21 L

<em>Note: the complete second part of the question is given below:</em>

<em>2. Suppose 0.470 kg of nonane are burned in air at a pressure of exactly 1 atm and a temperature of 17.0 °C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.</em>

Explanation:

Part 1: Balanced chemical equation

C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)

Part 2: volume of carbon dioxide produced

From the equation of the reaction;

At s.t.p., I mole of  C₉H₂₀ reacts with 14 moles of O₂ to produce 9 moles of CO₂

molar mass of  C₉H₂₀  = 128g/mol: molar mass of CO₂ = 44 g/mol, molar volume of gas at s.t.p. = 22.4 L

Therefore, 128 g of C₉H₂₀ produces 14 * 22.4 L of CO₂ i.e. 313.6 L of CO₂.

O.470 Kg  of nonane = 470 g of nonane

470 g of C₉H₂₀ will produce 470 * (313.6/128) L of CO₂ = 1151.50 L of CO₂

Volume of CO₂ gas produced at 1 atm and 17 °C;

Using P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂/P₂T₁

where P₁ = 1 atm, V₁ = 1151.50 L, T₁ = 273 K, P₂ = 1 atm, T₂ = 17 + 273 = 290 K

Substituting the values; V₂ = (1 * 1151.5 * 290)/(1 * 273)

Therefore volume of CO₂ produced, V₂ = 1223.21 L of CO₂

3 0
3 years ago
When 11.12 g of neon is combined in a 100 L container at 80oC with 9.59 g of argon, what is the mole fraction of argon?
PolarNik [594]
The correct answer for the question that is being presented above is this one: "<span>0.3."

Here it is how to solve.
M</span><span>olecular mass of Ar = 40
</span><span>Molecular mass of Ne = 20
</span><span>Number of moles of Ar = 9.59/40 = 0.239
</span><span>Number of moles of Ne = 11.12/20= 0.556
</span><span>Mole fraction of argon = 0.239/ ( 0.239 + 0.556) = 0.3</span><span>
</span>
3 0
3 years ago
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