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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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Answer:

\large \boxed{5.5 \times 10^{22}\text{ molecules of P$_{2}$O}_{5}}

Explanation:

You must calculate the moles of P₄O₁₀, convert to moles of P₂O₅,  then convert to molecules of P₂O₅.

1. Moles of P₄O₁₀

\text{Moles of P$_{4}$O}_{10} = \text{13 g P$_{4}$O}_{10} \times \dfrac{\text{1 mol P$_{4}$O}_{10}}{\text{283.89 g P$_{4}$O}_{10}} = \text{0.0458 mol P$_{4}$O}_{10}

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P₄O₁₀ ⟶ 2P₂O₅

The molar ratio is 2 mol P₂O₅:1 mol P₄O₁₀

\text{Moles of P$_{2}$O}_{5} = \text{0.0458 mol P$_{4}$O}_{10} \times \dfrac{\text{2 mol P$_{2}$O}_{5}}{\text{1 mol P$_{4}$O}_{10}} = \text{0.0916 mol P$_{2}$O}_{5}

3. Molecules of P₂O₅

\text{No. of molecules} = \text{0.0916 mol P$_{2}$O}_{5} \times \dfrac{6.022 \times 10^{23}\text{ molecules P$_{2}$O}_{5}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{5.5 \times 10^{22}}\textbf{ molecules P$_{2}$O}_{5}\\\text{There are $\large \boxed{\mathbf{5.5 \times 10^{22}}\textbf{ molecules of P$_{2}$O}_{5}}$}

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3 years ago
Explain how creativity can play a role in the construction of scientific questions and hypotheses.
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