Question

A container of gas is initially at 0.750 atm and 45 ˚c. what will the pressure be at 105 ˚c

Answer 1

Using Gay-Lussac's Law, the pressure of a gas increases with temperature. By calculating with initial conditions of 0.750 atm at 45 ℃ and converting to Kelvin, the final pressure is found to be approximately 0.890 atm at 105 ℃.

Step-by-step explanation:

The student's question pertains to the change in pressure of a gas when subjected to a change in temperature while the volume remains constant. To find the new pressure at 105 ℃ when the initial pressure is 0.750 atm at 45 ℃, we can use Gay-Lussac's Law which states that the pressure of a gas is directly proportional to its temperature in kelvins (P/T = k). First, we convert the temperatures from Celsius to Kelvin (K = ℃ + 273.15), then apply the formula to find the final pressure:

Initial temperature (T1) = 45 ℃ = 318.15 K
Initial pressure (P1) = 0.750 atm
Final temperature (T2) = 105 ℃ = 378.15 K

Using the formula: P2 = (P1 * T2) / T1

Now we calculate:
P2 = (0.750 atm * 378.15 K) / 318.15 K
P2 = 0.890 atm (approximately)

So the final pressure is 0.890 atm at 105 ℃.