Problem Statement:
16 kilograms of oxygen gas expands at STP (1 ATM) isobarically to occupy double its original value. What is the work done in the process?

Solution:

To find the work done in the process, we need to use the formula for work done in an isobaric process:

Work (W) = Pressure (P) * Change in Volume (ΔV)

Given:
- Initial mass of oxygen gas (m) = 16 kilograms
- Initial pressure (P1) = 1 ATM
- Final volume (V2) = 2 times the initial volume (V1)

Step 1: Calculate the initial volume (V1)

To calculate the initial volume, we can use the ideal gas law equation:

PV = nRT

Where:
- P is the pressure (1 ATM)
- V is the volume (unknown)
- n is the number of moles of gas (unknown)
- R is the ideal gas constant (0.0821 L·atm/(mol·K))
- T is the temperature in Kelvin (STP = 273.15 K)

Rearranging the equation, we get:

V = (nRT) / P

Since we know the mass of oxygen gas (16 kilograms) and the molar mass of oxygen (32 g/mol), we can calculate the number of moles (n) using the formula:

n = m / M

Where:
- m is the mass of the gas (16 kilograms)
- M is the molar mass of the gas (32 g/mol)

Converting the mass from kilograms to grams, we get:

m = 16,000 grams

Substituting the values into the equation, we can calculate the number of moles (n):

n = 16,000 grams / 32 g/mol = 500 moles

Now we can calculate the initial volume (V1) using the ideal gas law equation:

V1 = (nRT) / P1
V1 = (500 moles * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 ATM
V1 = 11,309.075 L

Therefore, the initial volume (V1) is 11,309.075 liters.

Step 2: Calculate the change in volume (ΔV)

The change in volume is given as double the initial volume:

ΔV = V2 - V1
ΔV = 2 * V1 - V1
ΔV = V1
ΔV = 11,309.075 L

Therefore, the change in volume (ΔV) is 11,309.075 liters.

Step 3: Calculate the work done (W)

Using the formula for work done in an isobaric process:

W = P * ΔV
W = 1 ATM * 11,309.075 L

Converting the units from ATM·L to kcal, we know that 1 ATM·L = 0.