18g of liquid water is placed in a flexible bag, the air is excluded, and the bag is sealed. It is then placed in a microwave oven where the water is boiled to make steam at 100C. What is the volume of the bag after all the water has boiled? Assume that the pressure inside the bag is equal to atmospheric pressure.
so this is what i did so far:
18 g of water = 1 mol.
then i used the ideal-gas equation,
PV = nRT, to find the volume at 100°C (= 373 K).
P = 1 atm
n = 1 mol
R = 0.083 L·atm/mol·K;
T = 373 K, so
V = nRT/P = 31 L
right?
yes. There was an easier way to do this. One mole of "steam" at STP is 22.4L
new volume: 22.4 * 373/273=you guessed it.
This constant pressure is called Charles' Law.
i'll use that for future
Your calculations are incorrect. Let's go through the process step by step.
First, we need to convert the mass of water into moles. To do this, we can use the molar mass of water, which is approximately 18 g/mol. So, 18 g of water corresponds to 1 mole.
Next, we need to convert the volume of the bag from moles to liters. We can use the ideal gas 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.
You correctly set the pressure as 1 atm and the number of moles as 1. However, you used the incorrect ideal gas constant. The correct value is 0.0821 L·atm/mol·K, not 0.083 L·atm/mol·K.
The temperature T should also be converted to Kelvin. You mentioned that the water is boiled at 100°C, which corresponds to 373 K (100°C + 273 K).
Using the correct values, we can calculate the volume V:
V = (nRT)/P
V = (1 mol) * (0.0821 L·atm/mol·K) * (373 K) / (1 atm)
V ≈ 30.43 L
Therefore, the volume of the bag after all the water has boiled is approximately 30.43 liters.