Correct answer is (C)

On P–V scale area of loop = work done

⇒ W = +1/2 (2) x 300

W = 300 J

Correct answer is (D)

When the gas is compressed suddenly, it undergoes an adiabatic process where no heat is exchanged with the surroundings.

Therefore, we can use the adiabatic equation of state to relate the initial and final pressure and volume of the gas: $P_0{V}_0^{\gamma }=P_f{V}_f^{\gamma }$ where $P_f$ and $V_f$ are the final pressure and volume of the gas, respectively.

Since the gas is compressed to $\frac{V_0}{4}$, we have: $V_f=\frac{V_0}{4}$ Substituting this into the adiabatic equation of state, we get: $P_f=P_0{\left(\frac{V_0}{V_f}\right)}^{\gamma }=P_0{\left(\frac{4}{1}\right)}^{\gamma }=4^{\gamma }{P}_0$

Therefore, the final pressure of the gas when it is suddenly compressed to $\frac{V_0}{4}$ is $4^{\gamma }{P}_0$.

Correct answer is (D)

If the temperature (T) remains constant, the internal energy (U) also remains constant, since the internal energy of an ideal gas depends only on its temperature.

In this case, the thermodynamic process in which the internal energy of the system remains constant is an isothermal process. Isothermal processes occur at constant temperature, and for an ideal gas, this means that the internal energy remains constant as well.

Correct answer is (C)

An adiabatic process is one in which there is no heat exchange between a system (in this case, the gas) and its surroundings. This happens because the system is perfectly insulated or the process occurs very quickly, not allowing for heat exchange.

Given the options:

Option A: There is no heat supplied to the system.

• This statement is TRUE. In an adiabatic process, there is no heat exchange between the system and its surroundings, as mentioned above.

Option B: The temperature of the gas increases.

• This statement is TRUE. When a gas is compressed adiabatically, the work done on the gas increases its internal energy, which in turn increases the temperature of the gas.

Option C: There is no change in the internal energy.

• This statement is NOT TRUE. In an adiabatic process, the change in internal energy of the system is equal to the work done on the system. When the gas is compressed, work is done on the gas, which increases its internal energy.

Option D: The change in the internal energy is equal to the work done on the gas.

• This statement is TRUE. As mentioned above, in an adiabatic process, the change in internal energy of the system is equal to the work done on the system.

So, Option C is the statement that is not true for an adiabatic process.

Correct answer is (C)

The final pressure of gas in container A after isothermal compression can be found using the equation of state for an ideal gas, $PV=nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature. For an isothermal process, the temperature $T$ is constant, so the equation becomes $P_1{V}_1=P_2{V}_2$.

The final volume is $\frac{1}{8}$ of the initial volume, so $P_2=P_1\times \frac{V_1}{V_2}=P_1\times 8=8P_1$.

The final pressure of gas in container B after adiabatic compression can be found using the adiabatic equation for an ideal gas, $P{V}^{\gamma }=\text{constant}$, where $\gamma$ is the ratio of the heat capacities, which is $\frac{5}{3}$ for a monoatomic gas.

Since $V_2=\frac{V_1}{8}$, we have $P_2=P_1\times {\left(\frac{V_1}{V_2}\right)}^{\gamma }=P_1\times 8^{\frac{5}{3}}=P_1\times 2^5=32P_1$.

The ratio of the final pressure of gas in B to that of gas in A is therefore $\frac{32P_1}{8P_1}=4$.