Detailed Answer

Adiabatic means there is no thermal energy exhanged between the system and the surroundings. This is only possible when the expansion happens very quickly, so there is simply no time for heat to be exchanged.

Detailed Answer

The first law of thermodynamics states:

$$Q = \Delta U + W$$

Here, the work done is positive, as it does work on the surroundings, and the heat absorbed is also positive. Substituting the values, we have:

$$\Delta U = Q - W = 80 - 50$$

This gives:

$$\Delta U = 30$$

The correct answer is D.

Detailed Answer

During an isothermal change, the temperature of the gas remains unchanged, hence the name. This means the change in interal energy is 0, so $Q = W$, and since its expanding the work is being done by the gas.

Detailed Answer

Internal energy is the measure of the temperature of the object, but in Kelvin. So for this problem we first need to convert the temperature. $20C = 20+273 = 293K$, which we can double, and convert back to °C to get D.

Detailed Answer

For this problem, we need to find the area under the graph. This can be done by splitting the graph into a rectangle and a triangle. The area of the rectangle is $2 \times 10^5$ and the area of the triangle above it is $\frac{1}{2} \times 2 \times 10^5$. Adding them gives $30,000 \, \text{J}$, which is $0.3 \, \text{MJ}$.

Detailed Answer

Efficiency is always the useful work divided by the total work. If $500 \, \text{J}$ come in, and $300 \, \text{J}$ go out, it means $200 \, \text{J}$ is the useful work done by the engine, so the efficiency is $\frac{200}{500}$, which is C.

Detailed Answer

The efficiency of a carnot engine is $1 - \frac{Q_{\text{cold}}}{Q_{\text{hot}}}$, replacing the letters gives D.

Detailed Answer

For an isothermal process, the change in internal energy ($\Delta U$) is zero, since the temperature is constant. According to the first law of thermodynamics: $$\Delta U = Q - W$$ Since the temperature is constant, $\Delta U = 0$. Therefore, the work done by the gas ($W$) is equal to the heat absorbed ($Q$).

Detailed Answer

The work done by a gas during an expansion at constant pressure is given by: $$W = P \Delta V$$ where: $P = 3.0 \, \text{Pa}$, $\Delta V = V_f - V_i = 4.0 \, \text{m}^3 - 2.0 \, \text{m}^3 = 2.0 \, \text{m}^3$. Substitute the values: $$W = (3.0)(2.0) = 6.0 \, \text{J}$$

Detailed Answer

According to the first law of thermodynamics: $$\Delta U = Q - W$$ where: $Q = 400 \, \text{J}$, $W = 150 \, \text{J}$. Substitute the values: $$\Delta U = 400 - 150 = 250 \, \text{J}$$

Detailed Answer

The change in entropy ($\Delta S$) during a reversible isothermal process is given by: $$\Delta S = \frac{Q}{T}$$ where: $Q = 500 \, \text{J}$, $T = 300 \, \text{K}$. Substitute the values: $$\Delta S = \frac{500}{300} = 1.67 \, \text{J/K}$$

Detailed Answer

The second law of thermodynamics states that the total entropy of an isolated system always increases over time.

Detailed Answer

In an adiabatic process, there is no heat exchange ($Q = 0$), and the internal energy changes only due to the work done.

Detailed Answer

The efficiency of a Carnot engine is given by the formula: $$\eta = 1 - \frac{T_C}{T_H}$$ where: $T_H = 600 \, \text{K}$ (hot reservoir), $T_C = 300 \, \text{K}$ (cold reservoir). Substitute the values: $$\eta = 1 - \frac{300}{600} = 1 - 0.5 = 0.5 = 50 \%$$