Question

A piston cylinder contains gas initially at 3500 kPa with a volume of 0.03 cubic meter. The gas is compressed during a process where pV raise to 1.25 = C to a pressure of 8500 kPa. The heat transfer from the gas is 2.5 kJ. Determine the change in internal energy, neglecting changes in kinetic and potential energies. Graph a PV and TS Plane.

Answer 1

Answer on Question #72721, Engineering / Mechanical Engineering

A piston cylinder contains gas initially at $3500\,\mathrm{kPa}$ with a volume of 0.03 cubic meter. The gas is compressed during a process where pV raise to $1.25 = C$ to a pressure of $8500\,\mathrm{kPa}$. The heat transfer from the gas is $2.5\,\mathrm{kJ}$. Determine the change in internal energy, neglecting changes in kinetic and potential energies. Graph a PV and TS Plane.

Solution:

The polytropic process is one in which the pressure-volume relation is given as

$$p V^n = \text{constant}$$

In our case,

$$p V^{1.25} = \text{constant}$$ $$p_1 V_1^{1.25} = p_2 V_2^{1.25}$$

Thus,

$$V_2 = V_1 \left(\frac{p_1}{p_2}\right)^{\frac{1}{1.25}} = 0.03 \left(\frac{3500}{8500}\right)^{\frac{1}{1.25}} = 0.01475\,\mathrm{m}^3$$

The work done during the polytropic process is found by substituting the pressure volume relation into the boundary work equation. The result is

$$W = \int_1^2 p\,dV = \frac{p_2 V_2 - p_1 V_1}{1 - n}$$

So,

$$W = \frac{8500 \times 10^3 \times 0.01475 - 3500 \times 10^3 \times 0.03}{1 - 1.25} = -81500\,\mathrm{J}$$

Then, according to the first law of thermodynamics,

$$\Delta U = Q - W$$

where $\Delta U$ is the change in the internal energy of the system and $W$ is work done by the system.

$$\Delta U = -2500 + 81500 = 79000\,\mathrm{J} = 79\,\mathrm{kJ}$$

Answer: $\Delta U = 79\,\mathrm{kJ}$

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