Question

A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m^3, undergoes a constant-pressure expansion at 2 bar to a final volume of 0.12 m^3, 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 atmosphere pressure is 1 bar.
a.) For the gas as the system, evaluate the work and the heat transfer, each in
b.) For the piston as a syatem, evautate the work and change in potential energy, in kJ.

Answer 1

Answer:

Q = E + W = 0.25 + 0.4 = 0.29KJ

W = 0.4KJ

b) 2000KJ

Explanation:

The concept applied here is the first law of thermodynamics ,i.e the equation of the first law.

mathematically from first law,

dE = dQ - dW

where E is internal energy

Where energy may be transfereed as eitherf work (W) or heat (Q)

Work on the other hand can be at constant volume and pressure.

at constant pressure, W = Integral (pdV), with final volume(Vf) as the upper limit and initial volume(Vi) as the lower limit

Work = p(Vf -Vi)

attempting the first question, Vi = 0.1metre cube, Vf = 0.12metre cube, p = 2bar, p(atm) = 1bar, E = U = 0.25KJ

from W = p(Vf -Vi) = 2 x 100000 ( 0.12 - 0.1) = 400N/m = 0.4KJ

To get the heat transfer, from dE = dQ - dW

Q = E + W = 0.25 + 0.4 = 0.29KJ

b) for the second question ; from Pdv = p(Vf -Vi), but the pressure here is in atmosphere

W = 100000 ( 0.12 - 0.1) = 2000KJ

Answer 2

Answer: (a). W = 4KJ and Q = 4.25KJ

(b). W = -2KJ and ΔPE = 2KJ

Explanation:

(a).

i. We are asked to calculate the work done during the expansion process considering gas as system.

from W = $\int\limits^a_b {p} \, dV$ where a = V₂ and b = V₁

so W = P(V₂-V₁)

W = (2 × 10²) (0.12 - 0.10)

W = 4 KJ

ii.  We apply the energy balance to gas as system

given Q - W = ΔE

Where ΔE = ΔU + ΔKE + ΔPE

since motion of the system is constrained, there is no change in both the potential and kinetic energy i.e. ΔPE = ΔKE = 0

∴ Q - W = ΔU

Q = ΔU + W

Q = 0.25 + 4

Q = 4.25 KJ

(b).

i. to calculate the work done during the expansion process considering piston as system;

W = $\int\limits^a_b {(Patm - Pgas)} \, dV$where a and b represent V₂ and V₁ respectively.

W = (Patm - Pgas)(V₂ - V₁)

W = (1-2) ×10² × (0.12-0.1)

W = -2KJ

ii. 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