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anygoal [31]
3 years ago
14

Suppose that 33.3 J of heat is added to an ideal gas. The gas expands at a constant pressure of 1.45 104 Pa while changing its v

olume from 3.24 10-4 to 8.40 10-4 m3. The gas is not monatomic, so the relation does not apply.(a) Determine the change in the internal energy of the gas.
Chemistry
1 answer:
levacccp [35]3 years ago
5 0

Answer:

ΔU = 25.8 J

Explanation:

The gas absorbs 33.3 J of heat, that is, Q = 33.3 J.

The work (W) of expansion can be calculated using the following expression:

W = -P. ΔV

where,

P is the external pressure

ΔV is the change in volume

W = -1.45 × 10⁴ N . m⁻² × (8.40 × 10⁻⁴ m³ - 3.24 × 10⁻⁴ m³) = -7.48 J

The change in the internal energy (ΔU) is:

ΔU = Q + W

ΔU =  33.3 J + (-7.48 J) = 25.8 J

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Answer:

Ksp = 8.8x10⁻⁵

Explanation:

<em>Full question is:</em>

<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>

<em />

When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:

PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻

Ksp = [Pb²⁺] [Cl⁻]²

If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):

Ksp = [X] [2X]²

Ksp = 4X³

As X is the amount of Pb²⁺ = 2.8x10⁻²M:

Ksp = 4(2.8x10⁻²)³

<h3>Ksp = 8.8x10⁻⁵</h3>
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