Question

Suppose you mix 4.0 mol of a monatomic gas at \(20.0^{\circ} \mathrm{C}\) and 3.0 mol of another monatomic gas at \(30.0^{\circ} \mathrm{C} .\) If the mixture is allowed to reach equilibrium, what is the final temperature of the mixture? [Hint: Use energy conservation.]

Short answer

Answer: The final equilibrium temperature of the mixture is approximately \(155^{\circ} \mathrm{C}\).

Step-by-step solution

Write the energy conservation equation

For energy conservation, the initial total internal energy equals the final total internal energy. So, we can write the equation as: $$U_{1} + U_{2} = U_{mix}$$

Write the internal energy formula for each gas

Since these are monatomic gases, we can use the formula \(U = \frac{3}{2}nRT\) for the internal energy. We can write the internal energy formula for both the gases and the mixture as follows: $$U_{1} = \frac{3}{2}n_{1}RT_{1}$$ $$U_{2} = \frac{3}{2}n_{2}RT_{2}$$ $$U_{mix} = \frac{3}{2}(n_{1}+n_{2})RT_{f}$$

Substitute the given values

Now, we can substitute the given values of the moles and temperatures into the energy conservation equation: $$\frac{3}{2}(4)(8.314)(20+273.15) + \frac{3}{2}(3)(8.314)(30+273.15) = \frac{3}{2}(4+3)(8.314)T_f$$

Solve for T_f

Now, we need to solve for the final temperature (T_f). First, simplify the equation: $$\frac{3}{2}(4)(8.314)(293.15) + \frac{3}{2}(3)(8.314)(303.15) = \frac{3}{2}(7)(8.314)T_f$$ $$14684.852+22609.709 = 87.105T_f$$ Continue by adding the left side and dividing by 87.105: $$37294.561 = 87.105T_f$$ $$T_f \approx 428.15 \mathrm{K}$$

Convert the final temperature into Celsius

To convert the final temperature from Kelvin to Celsius, we can use the following relation: $$T(\mathrm{^\circ C}) = T(\mathrm{K}) - 273.15$$ $$T_f(\mathrm{^\circ C}) \approx 428.15 - 273.15$$ $$T_f(\mathrm{^\circ C}) \approx 155^{\circ} \mathrm{C}$$ The final equilibrium temperature of the mixture is approximately \(155^{\circ} \mathrm{C}\).