Final answer:
The pressure at -25°C, with the volume kept constant, can be calculated using Gay-Lussac's Law. Given that the original conditions are STP with a pressure of 1 atm, the final pressure is found to be 0.91 atm after the temperature change.
Step-by-step explanation:
To calculate the pressure of a gas that was originally at standard temperature and pressure (STP) and then cooled to -25°C at constant volume, we can use Gay-Lussac's Law, which states that the pressure of a fixed amount of gas held at a constant volume is directly proportional to its temperature in kelvins. At STP, the pressure is 1 atm and the temperature is 0°C (273.15 K). To find the new pressure at -25°C (248.15 K), we need to set up the following ratio:
P1 / T1 = P2 / T2
where P1 is the initial pressure (1 atm), T1 is the initial temperature (273.15 K), P2 is the final pressure, and T2 is the final temperature (248.15 K).
By substituting the values and rearranging the equation to solve for P2, we get:
P2 = P1 * (T2 / T1) = 1 atm * (248.15 K / 273.15 K) = 0.908 atm, which rounds to 0.91 atm.
Therefore, the correct answer is c) 0.91 atm.