A rigid container holds a gas at a pressure of 55kPa and a temp of -100.0 deg C. What will the pressure be when the temp is increased to 200.0 deg C?

Answer 1

To solve this problem, we can use the Gay-Lussac's Law formula, which relates pressure and temperature of a gas in a rigid container:

P1/T1 = P2/T2

where P1 is initial pressure, T1 is initial temperature in Kelvin, P2 is final pressure, and T2 is final temperature in Kelvin.

First, let's convert the temperatures to Kelvin:
T1 = -100.0 + 273.15 = 173.15 K
T2 = 200.0 + 273.15 = 473.15 K

Now, plug the values into the equation and solve for P2:

55 kPa / 173.15 K = P2 / 473.15 K

P2 = (55 kPa * 473.15 K) / 173.15 K = 148.75 kPa

So, the final pressure will be approximately 148.75 kPa when the temperature is increased to 200.0°C.