Answer to Question #228383 in Molecular Physics | Thermodynamics for Unknown346307

Question

. The following equation gives the internal energy of a certain substance

u = 3.64 pv + 90

where u is kJ/kg, p is in kPa and v is in m3/kg.

A system composed of 3.5 kg of this substance expands from an initial pressure of 500 kPa

and a volume of 0.25 m3 to a final pressure 100 kPa in a process in which pressure and volume are

related by pv1.25 = constant.

(i) If the expansion is quasi-static, find Q, ∆U and W for the process.

(ii) In another process, the same system expands according to the same pressure-volume

relationship as in part (i), and from the same initial state to the same final state as in part (i), but

the heat transfer in this case is 32 kJ. Find the work transfer for this process.

(iii) Explain the difference in work transfer in parts (i) and (ii).

Answer 1

Gives

u=3.64pv+90

(1)

∆u=u₂-u₁

$$∆u=3.64(p_2V_2-p_1V_1)\\p_1V_1^{1.25}=p_2V_2^{1.25}\\V_2=V_1(\frac{p_1}{p_2})^{\frac{1}{1.25}}$$

$$V_2=0.25\times(\frac{500}{100})^{\frac{1}{1.25}}=0.906m^3$$

∆U=

$$3.64(100\times103\times0.906-500\times103\times0.25)$$

$$∆U=3.64(100\times105(0.906-5\times0.25)J$$

$∆U=-125.2KJ$

Quasi static process

$W=\smallint p.dV$

$W=\smallint p.dV=\frac{p_1V_1-p_2V_2}{n-1}$

$$W=\frac{500\times10^3\times0.25-100\times10^30.906}{1.25-1}=137.6KJ$$

$$Q=∆U+W=-125.2+137.6=12.4kJ$$

Q=12.4kJ

(2)

Here Q=32kJ

Since the end states are the same

∆U would remain the same as in(1)

W=Q-∆U=32-(-125.2)=157.2kJ

(3) the work in (2) is not equal to $\smallint pdV$ since the process is not quasi static