Ten moles of an ideal monoatomic gas, initially in state a at atmospheric pressure and temperature T_a = 27°C, is enclosed in a metal cylinder of volume V₀ fitted with a frictionless piston. The gas is suddenly compressed to state b with volume V₀/3. Now, keeping the piston stationary, the cylinder is submerged in a water bath of temperature 11°C until the gas reaches the temperature of the water bath, which is denoted as state c. Finally, while still in the water bath, the piston is brought slowly to its initial position, which is denoted as state f. If R is universal gas constant, then the correct option(s) is/are:
Given: 9^1/3 = 2.08

The correct options are:

1. The schematic P-V diagram of the processes described above is:

2. The change in internal energy in going from state a to b is 4860R.

3. The net change in the internal energy in the whole process is -240R.

Step-by-Step Explanation:

Given Data:

Number of moles of monoatomic gas, n = 10

Initial state (a): Pressure P_a = P₀ (1 atm), Temperature T_a = 27°C = 300 K, Volume V_a = V₀

For a monoatomic ideal gas, degrees of freedom f = 3, so its molar specific heat at constant volume is C_v = (3/2)R, and the adiabatic index is γ = C_p/C_v = 5/3.

Process 1: Sudden compression from state a to state b

Since the gas is compressed suddenly, no heat exchange occurs during this step, making process a → b adiabatic.

Volume at state b is V_b = V₀/3.

Using the adiabatic relation between temperature and volume, T V^γ-1 = constant:

$T_b{V}_b^{\gamma -1}=T_a{V}_a^{\gamma -1}$

$T_b=T_a{\left(\frac{V_a}{V_b}\right)}^{\gamma -1}=300\times {(\frac{V_0}{V_0/3}}^{\mathrm{5;}/3-1}=300\times 3^{2/3}$

Using 3^2/3 = (9)^1/3 = 2.08:

$T_b=300\times 2.08=624\text{K}$

Now, let's calculate the change in internal energy (ΔU_a→b) from state a to b:

$\Delta U_{a\to b}=n{C}_v(T_b-T_a)$

$\Delta U_{a\to b}=10\times \frac{3}{2}R\times (624-300)$

$\Delta U_{a\to b}=15R\times 324=4860R$

Thus, the change in internal energy from state a to state b is indeed 4860R.

Process 2: Cooling at constant volume from state b to state c

The piston is kept stationary, so the volume remains constant at V_c = V₀/3 (Isochoric process).

The cylinder is submerged in a water bath of temperature T_c = 11°C = 284 K until thermal equilibrium is established.

Process 3: Slow expansion from state c to state f

While remaining in the water bath, the piston is brought back slowly to its initial position V_f = V₀. Since the process is carried out slowly in a constant temperature water bath, process c → f is isothermal at T_f = T_c = 284 K.

Net Change in Internal Energy for the Whole Process (a → b → c → f):

The initial temperature of the gas is T_a = 300 K and the final temperature of the gas is T_f = 284 K.

Since internal energy of an ideal gas depends only on temperature:

$\Delta U_{\text{net}}=n{C}_v(T_f-T_a)$

$\Delta U_{\text{net}}=10\times \frac{3}{2}R\times (284-300)$

$\Delta U_{\text{net}}=15R\times (-16)=-240R$

Thus, the net change in internal energy across the whole process is -240R.

Analysis of the P-V Diagram:

1. a → b: Adiabatic compression where volume decreases from V₀ to V₀/3, and pressure increases steeply along an adiabatic curve.

2. b → c: Isochoric cooling at volume V₀/3, shown as a vertical downward line as pressure decreases due to cooling from 624 K to 284 K.

3. c → f: Isothermal expansion back to volume V₀ along an isotherm corresponding to 284 K.

This matches the provided P-V curve diagram perfectly.