Question

A 20.0 -\,L volume of an ideal gas in a cylinder with a piston is at a pressure of 3.2 atm. Enough weight is suddenly removed from the piston to lower the external pressure to 1.6 atm. The gas then expands at constant temperature until its pressure is 1.6 atm.
1. Find Delta E for this change in state in J
2. Find Delta H for this change in state in J
3. Find q for this change in state in J
4. Find w for this change in state in J

Answer 1

Answer:

1. ΔE = 0 J

2. ΔH = 0 J

3. q = 3.2 × 10³ J

4. w = -3.2 × 10³ J

Explanation:

The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.

The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.

P₁ × V₁ = P₂ × V₂

3.2 atm × 20.0 L = 1.6 atm × V₂

V₂ = 40 L

The work (w) can be calculated using the following expression.

w = - P . ΔV

where,

P is the external pressure for which the process happened

ΔV is the change in the volume

w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J

The change in the internal energy is:

ΔE = q + w

0 = q + w

q = - w = 3.2 × 10³ J