a 5.00 Liter tank holds argon gas a 675 kPa and 11C. what would be its pressure if this argon is transferred to a different tanks whose volume is 17.0 liters at 135C?

(P1V1/T1) = (P2V2/T2)

Don't forget T is in Kelvin.

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

(P1 * V1) / (T1) = (P2 * V2) / (T2)

Where:
P1 = initial pressure of the first tank in kPa
V1 = initial volume of the first tank in liters
T1 = initial temperature of the first tank in Kelvin
P2 = final pressure of the second tank (what we are trying to find) in kPa
V2 = final volume of the second tank in liters
T2 = final temperature of the second tank in Kelvin

Now let's plug in the given values:

P1 = 675 kPa
V1 = 5.00 liters
T1 = 11°C + 273.15 = 284.15 K (converting Celsius to Kelvin)
V2 = 17.0 liters
T2 = 135°C + 273.15 = 408.15 K

Substituting these values into the combined gas law equation, we get:

(675 kPa * 5.00 L) / 284.15 K = (P2 * 17.0 L) / 408.15 K

Now let's solve for P2:

P2 = (675 kPa * 5.00 L * 408.15 K) / (17.0 L * 284.15 K)

P2 ≈ 1533.44 kPa

Therefore, the pressure of the argon gas in the second tank would be approximately 1533.44 kPa.