A scuba diver's tank contains 0.23 kg of O2 compressed into a volume of 3.5 L.

(a) Calculate the gas pressure inside the tank at 11°C.
________atm

(b) What volume would this oxygen occupy at 20°C and 0.92 atm?
________L

what are the formulas for this problem?

Use PV = nRT for 1.

Use (P1V2)/T1 = (P2V2/T2) for 2.

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To solve this problem, we will use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.

For part (a), we need to calculate the gas pressure inside the tank at 11°C. To do this, we can use the ideal gas law equation.

Step 1: Convert the temperature from Celsius to Kelvin.
T(K) = T(°C) + 273.15
T(K) = 11°C + 273.15
T(K) = 284.15 K

Step 2: Calculate the number of moles of oxygen.
Given mass of oxygen = 0.23 kg
Molar mass of oxygen (O2) = 32 g/mol
Number of moles (n) = mass/molar mass
n = (0.23 kg * 1000 g/kg) / 32 g/mol
n ≈ 7.1875 mol (round to four decimal places)

Step 3: Substitute the values into the ideal gas law equation and solve for pressure (P).
P1 * V1 = n * R * T
P1 = (n * R * T) / V1

Substituting the values:
P1 = (7.1875 mol * 0.0821 L·atm/(mol·K) * 284.15 K) / 3.5 L
P1 ≈ 14.629 atm (round to three decimal places)

Therefore, the gas pressure inside the tank at 11°C is approximately 14.629 atm.

For part (b), we need to find the volume that the oxygen would occupy at 20°C and 0.92 atm.

Step 1: Convert the temperature from Celsius to Kelvin.
T(K) = T(°C) + 273.15
T(K) = 20°C + 273.15
T(K) = 293.15 K

Step 2: Solve for V2 using the ideal gas law equation.
P1 * V1 = P2 * V2
V2 = (P1 * V1) / P2

Substituting the given values:
V2 = (14.629 atm * 3.5 L) / 0.92 atm
V2 ≈ 55.568 L (round to three decimal places)

Therefore, at 20°C and 0.92 atm, the oxygen would occupy approximately 55.568 L volume.