Question

A 20. 0-L stainless steel container at 25°C was charged with

2. 00 atm of hydrogen gas and 3. 00 atm of oxygen gas. A spark

ignited the mixture, producing water. What is the pressure in

the tank at 25°C? If the exact same experiment were per-

formed, but the temperature was 125°C instead of 25°C, what

would be the pressure in the tank?

Answer 1

To solve this problem, we can use the combined gas law, which relates the initial and final conditions of a gas sample. The combined gas law equation is:

(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

Let's solve the problem step by step:

Step 1: Convert the temperatures to Kelvin.

Given: T1 = 25°C, T2 = 125°C

Convert them to Kelvin:

T1 = 25°C + 273.15 = 298.15 K

T2 = 125°C + 273.15 = 398.15 K

Step 2: Determine the initial and final volumes.

Given: V1 = V2 = 20.0 L

Step 3: Substitute the values into the combined gas law equation.

For the initial condition (P1, V1, T1):

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

For the final condition (P2, V2, T2):

(P2 * V2) / T2 = (P2 * 20.0) / 398.15

Step 4: Solve for P2.

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

For the initial condition:

P1 = 2.00 atm (hydrogen)

V1 = 20.0 L

T1 = 298.15 K

Substituting these values:

P2 = (2.00 * 20.0 * 398.15) / (20.0 * 298.15) = 2.67 atm

So, at 25°C, the pressure in the tank is approximately 2.67 atm.

For the final condition (T2 = 398.15 K):

P2 = (2.00 * 20.0 * 398.15) / (20.0 * 398.15) = 2.00 atm

So, at 125°C, the pressure in the tank is 2.00 atm.

Please note that the volume remains constant in this problem, so it doesn't affect the pressure calculations.