One gas of $$n_1$$ mole of molecules at temperature $$T_1$$, volume $$V_1$$, and pressure $$P_1$$, and another gas of $$n_2$$ mole of molecules at temperature $$T_2$$, volume $$V_2$$, and pressure $$P_2$$, are mixed resulting in pressure $$P$$ and volume $$V$$ of the mixture. The temperature of the mixture is :

Assume both gases behave ideally. For an ideal gas the relation is $$PV = nRT$$, where $$n$$ is the number of moles and $$R$$ is the universal gas constant.

For gas 1 before mixing,

$$P_1V_1 = n_1RT_1 \; \Rightarrow \; n_1 = \frac{P_1V_1}{RT_1} \; -(1)$$

For gas 2 before mixing,

$$P_2V_2 = n_2RT_2 \; \Rightarrow \; n_2 = \frac{P_2V_2}{RT_2} \; -(2)$$

After mixing the two samples, the total number of moles is the sum of $$n_1$$ and $$n_2$$:

$$n = n_1 + n_2 = \frac{P_1V_1}{RT_1} + \frac{P_2V_2}{RT_2} \; -(3)$$

The mixture finally occupies volume $$V$$ at pressure $$P$$ and temperature $$T$$, so once again using the ideal-gas equation,

$$PV = nRT \; -(4)$$

Insert $$n$$ from $$(3)$$ into $$(4)$$:

$$PV = \left( \frac{P_1V_1}{RT_1} + \frac{P_2V_2}{RT_2} \right) RT$$

The gas constant $$R$$ cancels from numerator and denominator, giving

$$PV = T\left( \frac{P_1V_1}{T_1} + \frac{P_2V_2}{T_2} \right) \; -(5)$$

Solve $$(5)$$ for the final temperature $$T$$:

$$T = \frac{PV}{\dfrac{P_1V_1}{T_1} + \dfrac{P_2V_2}{T_2}}$$

Multiply numerator and denominator by $$T_1T_2$$ to simplify:

$$T = \frac{T_1T_2PV}{T_2P_1V_1 + T_1P_2V_2}$$

Thus, the temperature of the mixture is

$$\boxed{\displaystyle T = \frac{T_1T_2PV}{T_2P_1V_1 + T_1P_2V_2}}$$

Option B is correct.