The volume of an ideal gas decreases by approximately 30.72% when the temperature is lowered from 100°C to 50°C and the pressure is increased by 25%.
The problem involves the relationship between volume, temperature, and pressure of an ideal gas, which is described by the ideal gas law. According to this law (PV = nRT), the volume of an ideal gas is directly proportional to its temperature (in kelvin) and inversely proportional to its pressure.
Here, we are given that the temperature of a gas is lowered from 100°C to 50°C and the pressure is increased by 25%. Because the temperature change will be calculated in kelvin, we must first convert the temperatures from °C to K by adding 273.
Thus, the initial temperature (T1) is 373K and the final temperature (T2) is 323K.
Let the initial volume be V1 and the final volume be V2. The initial pressure P1 increased by 25% results in the final pressure P2 = 1.25P1.
We use the combined gas law, which gives us the relation (P1V1)/T1 = (P2V2)/T2.
Substituting P2 = 1.25P1 and rearranging for V2, we get:
V2 = (V1 * T2 * P1)/(T1 * 1.25 * P1).
The pressures cancel out, leaving V2 = V1 * T2 / (1.25 * T1).
Plugging in the temperatures in kelvin, we have:
V2 = V1 * 323K / (1.25 * 373K).
Simplifying, V2 = V1 * 323 / 466.25.
Therefore, the new volume V2 is approximately 0.6928 times the initial volume V1. This represents a decrease of about 30.72% in the volume, since 1 - 0.6928 = 0.3072.