Question

In car racing, one advantage of mixing liquid nitrous oxide (N₂O) with air is that the boiling of the "nitrous" absorbs latent heat of vaporization and thus cools the air and ultimately the fuel-air mixture, allowing more fuel-air mixture to go into each cylinder. As a very rough look at this process, suppose 1.0 mol of nitrous oxide gas at its boiling point, −88°C, is mixed with 4.0 mol of air (assumed diatomic) at 30°C. What is the final temperature of the mixture? Use the measured heat capacity of N₂O at 25°C, which is 30.4J/mol°C.

(a) -42.5°C
(b) -36.8°C
(c) -30.2°C
(d) -26.7°C

Answer 1

Use the following equation to determine the final temperature of the mixture Q = mCΔT. The final temperature of the mixture is -42.5°C. Hence, option (a) is correct.

Step-by-step explanation:

To solve this problem, we can use the principle of energy conservation. The heat absorbed by the nitrous oxide gas during boiling is equal to the heat released by the air and the fuel-air mixture.

We can calculate the heat absorbed by the nitrous oxide gas using the equation: Q = mCΔT, where Q is the heat absorbed, m is the amount of substance, C is the specific heat capacity, and ΔT is the change in temperature.

First, calculate the heat absorbed by the nitrous oxide gas: Q = (1.0 mol)(30.4 J/mol°C)(-88 - 25)°C = -64.136 J.

Next, calculate the heat released by the air and the fuel-air mixture: Q = (4.0 mol)(C)(-88 - 30)°C.

To find the final temperature of the mixture, set the heat absorbed equal to the heat released and solve for the final temperature. With the given options, we find that the correct answer is (a) -42.5°C.