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otez555 [7]
3 years ago
9

Oxygen gas is compressed in a piston–cylinder device from an initial state of 0.8 m3 /kg and 25°C to a final state of 0.1 m3 /kg

and 287°C. Determine the entropy change of the oxygen during this process. Assume constant specific heats.
Chemistry
1 answer:
zhuklara [117]3 years ago
3 0

Answer:

ΔS = -0.1076 kJ /kg*K

Explanation:

Step 1: Data given

Initial state = 0.8 m³/kg and 25 °C = 298.15 K

Final state = 0. 3³/kg and 287 °C = 560.15 K

Cv = 0.686 kJ/kg*K

Step 2: Calculate the average temperature

The average temperature = (25°C + 287 °C)/2  =156 °C ( = 429 K)

Step 3: Calculate the ΔS

ΔS =(Cv, average) * ln(T2/T1) + R*ln(V2/V1)

ΔS = 0.686 * ln(560.15/298.15) + 0.2598*ln( 0.1/0.8)

ΔS = -0.1076 kJ /kg*K

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9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8
MissTica

Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

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