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anygoal [31]
4 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]4 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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The triple point of nitrogen occurs at a temperature of 63.1 K and a pressure of 0.127 atm. Its normal boiling point is 77.4 K.
yulyashka [42]

Nitrogen has a normal boiling point of 77.4 K and a melting point (at 1 atm) of 63.2 K. Its critical temperature is 126.2 K, and its critical pressure is 2.55 * 104 torr. It has a triple point at 63.1 K and 94.0 torr.

<h3>What is the triple point?</h3>

The temperature and pressure at which the solid, liquid, and vapour phases of a pure substance can coexist in equilibrium.

- Normal melting point: 63.2 K.

- Normal boiling point: 77.4 K.

- Triple point: 0.127 atm and 63.1 K.

- Critical point: 33.5 atm and 126.0 K.

In such a way:

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- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.

- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.

- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.

Learn more about triple point here:

brainly.com/question/23307017

#SPJ1

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