Question

Consider a container with a frictionless piston that contains a given amount of an ideal gas. if the external pressure is kept constant, the piston will move up or down in response to a change in the internal pressure. the piston will move up if pint > pext and vice versa. the piston will stop moving when pint = pext (the system is equilibrated).

Answer 1

Since the problem is incomplete, I tried to search a similar question in the internet. Luckily, I found the exact problem which is shown in the attached picture. Here are the solution for the five questions:

1. Assuming ideal behavior,

PV = nRT
To use SI units, convert bar to Pa using the conversion: 1 bar = 100,000 Pa. Then, R = 8.314 m³·Pa/mol·K.
(1.15 bar)(100,000 Pa/1 bar)(V) = (0.9 mol)(8.314 m³·Pa/mol·K)(32 + 273 K)
Solving for V,
V = 0.0198 m³

2. Since the Pint=Pext because it is allowed to reach equilibrium, P is still 1.15 bar, but T is now 347°C.

PV = nRT
(1.15 bar)(100,000 Pa/1 bar)(V) = (0.9 mol)(8.314 m³·Pa/mol·K)(347 + 273 K)
Solving for V,
V = 0.0403 m³

3. W = PΔV
    W = (1.15 bar)(100,000 Pa/1 bar)(0.0403 m³ - 0.0198 m³)
    W = 2,357.5 J or 2.36 kJ

5. Let's go first with #5 because this is needed to solve for #4. Internal energy is ΔU.

ΔU = nCvΔT = (0.9 mol)(3.3)(8.314 m³·Pa/mol·K)(347°C - 32°C)
ΔU = 7,778.16 J or 7.78 kJ

4. The formula for Q is:
ΔU = Q + W
7.78 kJ = Q + 2.36 kJ
Solving or Q,
Q = 5.42 kJ