Oxygen (molar mass 32kg/kmol) is compressed reversible and poly tropically in the cylinder from 1.05 bar to 4.2 bar in such a way that one-third of work input is rejected as heat to the cylinder walls. Calculate the final temperature of oxygen. Assume oxygen to be a perfect gas and take Cv = 0.649

The molar mass of oxygen ($M$) is $32 \text{ kg/kmol}$. The universal gas constant ($R_u$) is $8.314 \text{ kJ/(kmol}\cdot\text{K)}$.

$$R = \frac{R_u}{M} = \frac{8.314 \text{ kJ/(kmol}\cdot\text{K)}}{32 \text{ kg/kmol}} = 0.2598125 \text{ kJ/(kg}\cdot\text{K)}$$

Given $C_v = 0.649 \text{ kJ/(kg}\cdot\text{K)}$. For a perfect gas, $C_p - C_v = R$.

$$C_p = C_v + R = 0.649 \text{ kJ/(kg}\cdot\text{K)} + 0.2598125 \text{ kJ/(kg}\cdot\text{K)} = 0.9088125 \text{ kJ/(kg}\cdot\text{K)}$$

The adiabatic index $k$ is:

$$k = \frac{C_p}{C_v} = \frac{0.9088125 \text{ kJ/(kg}\cdot\text{K)}}{0.649 \text{ kJ/(kg}\cdot\text{K)}} = 1.4003274$$

The first law of thermodynamics for a closed system is $\Delta U = Q + W_{on}$, where $W_{on}$ is work done on the system and $Q$ is heat added to the system.
Work input ($W_{in}$) is done on the system, so $W_{on} = W_{in}$.
Heat rejected ($Q_{rej}$) means heat leaves the system, so $Q = -Q_{rej}$.
The problem states that "one-third of work input is rejected as heat", so $Q_{rej} = \frac{1}{3} W_{in}$.
Therefore, $Q = -\frac{1}{3} W_{in}$.
Substituting into the first law:

$$\Delta U = -\frac{1}{3} W_{in} + W_{in} = \frac{2}{3} W_{in}$$

For a perfect gas, $\Delta U = m C_v (T_2 - T_1)$. For a polytropic process, work input $W_{in} = \frac{m R (T_2 - T_1)}{n-1}$.
Equating these expressions:

$$m C_v (T_2 - T_1) = \frac{2}{3} \frac{m R (T_2 - T_1)}{n-1}$$

Assuming $T_2 \neq T_1$, we can simplify to:

$$C_v = \frac{2}{3} \frac{R}{n-1}$$

Rearranging to solve for $n$:

$$n-1 = \frac{2R}{3C_v}$$ $$n = 1 + \frac{2R}{3C_v} = 1 + \frac{2 \times 0.2598125 \text{ kJ/(kg}\cdot\text{K)}}{3 \times 0.649 \text{ kJ/(kg}\cdot\text{K)}} = 1 + \frac{0.519625}{1.947} = 1 + 0.2668849512 \approx 1.2669$$

The problem does not provide an initial temperature. We will assume a standard room temperature of $25^\circ\text{C}$.

$$T_1 = 25^\circ\text{C} = 25 + 273.15 = 298.15 \text{ K}$$

For a polytropic process, the relation between temperature and pressure is:

$$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}}$$

Given $P_1 = 1.05 \text{ bar}$ and $P_2 = 4.2 \text{ bar}$.
First, calculate the exponent $\frac{n-1}{n}$:

$$\frac{n-1}{n} = \frac{0.2668849512}{1.2668849512} = 0.2106504854$$

Now, substitute the values into the temperature-pressure relation:

$$\frac{T_2}{298.15 \text{ K}} = \left(\frac{4.2 \text{ bar}}{1.05 \text{ bar}}\right)^{0.2106504854}$$ $$\frac{T_2}{298.15 \text{ K}} = (4)^{0.2106504854}$$ $$(4)^{0.2106504854} \approx 1.339106$$ $$T_2 = 298.15 \text{ K} \times 1.339106$$ $$T_2 \approx 399.243 \text{ K}$$

The final temperature of oxygen is approximately $\text{399.24 K}$.