Final answer:
The new temperature of the gas, after the pressure increase from 0.96 atm to 1.25 atm in a rigid container, is found to be 387.7 K using Gay-Lussac's law.
Step-by-step explanation:
A student asked: "A gas in a rigid container at 25°C has a pressure of 0.96 atm. A change in temperature causes the pressure to increase to 1.25 atm. What is the new temperature of the gas?" To answer this, we can use the ideal gas law in its combined form, known as Gay-Lussac's law, which states that for a given mass and constant volume of an ideal gas, the pressure is directly proportional to its temperature in Kelvin.
First, we need to convert the initial temperature from °C to K by adding 273.15: T1 = 25 + 273.15 = 298.15 K. Then we set up the proportionality:
(P1 / T1) = (P2 / T2)
Now we can solve for the new temperature T2 by rearranging the formula:
T2 = T1 * (P2 / P1) = 298.15 K * (1.25 atm / 0.96 atm)
T2 = 387.7 K
Thus, the new temperature of the gas is 387.7 K.