A canister holds a gas at 16.3 psi when the temperature is 15˚C. To what will the pressure change when the temperature is increased to 20˚C? *

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To answer this question, we can use the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature when the volume and the number of gas molecules remain constant. The ideal gas law equation is given as:

PV = nRT

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

Since we are given the pressure in psi and the temperature in ˚C, we need to convert these units before using the ideal gas law.

1. First, let's convert the initial temperature of 15˚C to Kelvin:
We know that 0˚C = 273.15 K, so we add 273.15 to 15˚C to convert it to Kelvin:
T1 = 15˚C + 273.15 = 288.15 K

2. Next, let's convert the pressure from psi to the appropriate unit for the ideal gas law, which is pascal (Pa):
1 psi = 6894.76 Pa
P1 = 16.3 psi × 6894.76 Pa/psi ≈ 112,446.7 Pa

Now we have the initial temperature (T1 = 288.15 K) and the initial pressure (P1 = 112,446.7 Pa). We can rearrange the ideal gas law equation to solve for the final pressure (P2) when the temperature increases to 20˚C:

P1/T1 = P2/T2

3. Convert the final temperature of 20˚C to Kelvin:
T2 = 20˚C + 273.15 = 293.15 K

4. Plug the values into the equation and solve for P2:
P2 = P1 × T2/T1
P2 = 112,446.7 Pa × 293.15 K / 288.15 K

Calculating this expression gives us the final pressure in pascals (Pa) when the temperature is increased to 20˚C.