Question

for part a) got W = 4kJ, dQ = 4.25 kJ
for part b) got w = 2kJ...

How do you find ΔPE without mass? Do you use the conservation of energy equation?

As shown in Fig. P2.56, a gas contained within a piston–
cylinder assembly, initially at a volume of 0.1 m3, undergoes
a constant-pressure expansion at 2 bar to a final volume of
0.12 m3, while being slowly heated through the base. The
change in internal energy of the gas is 0.25 kJ. The piston
and cylinder walls are fabricated from heat-resistant material,
and the piston moves smoothly in the cylinder. The local
atmospheric pressure is 1 bar.

a) For the gas as the system, evaluate work and heat transfer, each in kJ.

b) For the piston as the system, evaluate work and change in potential energy, each in kJ.

Answer 1

help

Answer 2

So you know how to get the work.

W = (Patm - Pgas)(V₂ - V₁)

W = (1-2) ×10² × (0.12-0.1)

W = -2KJ

We never have to use the mass. We apply the energy balance to gas as system

given Q - W = ΔE

Where ΔE = ΔU + ΔKE + ΔPE

Q = 0 since the piston and cylinder walls are perfectly insulated.

for piston, we neglect the change in internal energy and kinetic energy

ΔU = ΔKE = 0

from Q - W = ΔU + ΔKE + ΔPE

0 - (-2) = 0 + 0 + ΔPE

∴ ΔPE = 2KJ

Answer 3

$$ \begin{aligned} &\text { Using the change in total energy of a system }\\ &\begin{array}{l} \Delta E=\Delta U+\Delta K E+\Delta P E \\ d Q-W=U_{2}-U_{1}+0+\Delta P E \\ \begin{aligned} \Delta P E &=\left(U_{2}-U_{1}\right)-d Q+W \\ &=0.25-4.25+4 \\ &=0 \mathrm{~kJ} \\ \Delta P E &=0 \mathrm{~kJ} \end{aligned} \end{array} \end{aligned} $$