A rigid container of O2 has a pressure of 340 kPa at a temperature of 713 K. What is the pressure at 273 K?

P1/T1 = P2/T2

Don't forget to change T to Kelvin.

Thank you for both!!

To solve this problem, we can use the combined gas law equation, which states:

(P1 × V1) / (T1) = (P2 × V2) / (T2)

Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature

In this case, the initial pressure (P1) is 340 kPa, the initial temperature (T1) is 713 K, and the final temperature (T2) is 273 K (since 273 K is the given temperature).

Since the volume (V) is constant for a rigid container, we can cancel out V1 and V2 from the equation.

Now we can rearrange the equation to solve for P2:

P2 = (P1 × T2) / (T1)

Plugging in the given values:

P2 = (340 kPa × 273 K) / (713 K)

Calculating this equation:

P2 ≈ 130.08 kPa

Therefore, the pressure at 273 K is approximately 130.08 kPa.

To derive the pressure at a different temperature, we can use the ideal gas law equation:

PV = nRT

where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature in Kelvin

In this case, we have a rigid container, which means the volume (V) remains constant. Since we are interested in finding the pressure (P) at a different temperature (T), we can rewrite the equation as:

P1/T1 = P2/T2

P1 = initial pressure
T1 = initial temperature
P2 = final pressure
T2 = final temperature

Now we can plug in the given values:
P1 = 340 kPa
T1 = 713 K
T2 = 273 K (given)

Plugging these values into the equation, we get:

340 kPa / 713 K = P2 / 273 K

Now, we can rearrange the equation to solve for P2:

P2 = (340 kPa / 713 K) * 273 K

Evaluating this expression, we find:

P2 ≈ 130.87 kPa

So, the pressure of the rigid container of O2 at 273 K would be approximately 130.87 kPa.