A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure $P_0$, volume $V_0$, and temperature $T_0$. If the gas mixture is adiabatically compressed to a volume $V_0/4$, then the correct statement(s) is/are. (Given $2^{1.2}=2.3$; $2^{3.2}=9.2$; $R$ is gas constant)

Step 1: Calculate the molar heat capacity and adiabatic exponent of the mixture.

The molar heat capacities at constant volume for monatomic and rigid diatomic gases are:

$$C_{V,1} = \frac{3}{2}R$$ $$C_{V,2} = \frac{5}{2}R$$

The molar heat capacity of the mixture is:

$$C_{V,\text{mix}} = \frac{n_1 C_{V,1} + n_2 C_{V,2}}{n_1 + n_2}$$

Substitute $n_1 = 5$ and $n_2 = 1$:

$$C_{V,\text{mix}} = \frac{5\left(\frac{3}{2}R\right) + 1\left(\frac{5}{2}R\right)}{5 + 1}$$ $$C_{V,\text{mix}} = \frac{5}{3}R$$

The adiabatic exponent of the mixture is:

$$\gamma = 1 + \frac{R}{C_{V,\text{mix}}}$$ $$\gamma = 1 + \frac{3}{5}$$ $$\gamma = 1.6$$

Step 2: Determine the final temperature and pressure.

For an adiabatic process, the temperature-volume and pressure-volume relations are:

$$T_f = T_0 \left(\frac{V_0}{V_f}\right)^{\gamma-1}$$ $$P_f = P_0 \left(\frac{V_0}{V_f}\right)^{\gamma}$$

Substitute $V_f = \frac{V_0}{4}$ and $\gamma = 1.6$:

$$T_f = T_0 (4)^{0.6}$$ $$T_f = T_0 (2)^{1.2}$$ $$T_f = 2.3 T_0$$

Now calculate the final pressure:

$$P_f = P_0 (4)^{1.6}$$ $$P_f = P_0 (2)^{3.2}$$ $$P_f = 9.2 P_0$$

Since $9 P_0 < P_f < 10 P_0$, the statement about the final pressure is correct.

Step 3: Calculate the work done on the gas.

For an adiabatic process, the work done on the gas equals the change in internal energy:

$$W_{\text{on}} = \Delta U$$ $$W_{\text{on}} = n C_{V,\text{mix}} (T_f - T_0)$$

Substitute $n = 6$, $C_{V,\text{mix}} = \frac{5}{3}R$, and the temperatures:

$$W_{\text{on}} = 6 \left(\frac{5}{3}R\right) (2.3 T_0 - T_0)$$ $$W_{\text{on}} = 10 R (1.3 T_0)$$ $$W_{\text{on}} = 13 R T_0$$

Step 4: Calculate the final internal energy.

The final internal energy of the mixture is:

$$U_f = n C_{V,\text{mix}} T_f$$ $$U_f = 6 \left(\frac{5}{3}R\right) (2.3 T_0)$$ $$U_f = 23 R T_0$$

Since $23 R T_0$ is not in the range $(18 R T_0, 19 R T_0)$, the statement regarding the final internal energy is incorrect.

Answer: The correct statements are:

• The work done on the gas is $13 R T_0$.
• The final pressure $P_f$ of the gas is between $9 P_0$ and $10 P_0$.
• The adiabatic constant $\gamma$ of the mixture is $1.6$.