Question

A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3 . The propane undergoes a process to a final pressure of 4 bar, during which the pressure–volume relationship is pV1.1 = constant. For the propane, evaluate the work and heat transfer, each in kJ. Kinetic and potential energy effects can be ignored.

Answer 1

Answer:

Heat transfer in the process is -19.6 kJ

Work done in the process is - 28 kJ

Explanation:

As we know that the process equation is given as

$P_1V_1^{1.1} = P_2V_2^{1.1}$

now we know that

$P_1 = 1 bar$

$V_1 = 0.2 m^3$

$P_2 = 4 bar$

now from above equation we have

$1(0.2)^{1.1} = 4(V)^{1.1}$

$V = 0.057 m^3$

now work done in this process is given as

$W = \frac{P_1V_1 - P_2V_2}{N - 1}$

here we know N = 1.1

so we have

$W = \frac{10^5 (0.2) - 4 \times 10^5(0.057)}{1.1 - 1}$

$W = -2.8 \times 10^4 J$

$W = -28 KJ$

Now we have

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

$T_2 = \frac{P_2V_2 T_1}{P_1V_1}$

$T_2 = \frac{4\times 10^5 (0.057) (300)}{10^5 (0.2)}$

$T_2 = 342 K$

now we have

$\Delta U = \frac{f}{2} n R(\Delta T)$

$\Delta U = 3(P_2 V_2 - P_1 V_1)$

$\Delta U = 3(4 \times 10^5(0.057) - 10^5(0.2))$

$\Delta U = 8400$

so we have

$Q = \Delta U + W$

$Q = -28 kJ + 8.4 kJ$

$Q = -19.6 kJ$