In a fit of boredom, a chemistry teaching
assistant swallows a small piece of solid dry ice (CO2). The 8.9 g of dry ice are converted very quickly into gas at 311 K. Assuming the TA’s stomach has a volume of 1.5 L, what pressure is exerted on his stomach?
n = grams/molar mass
Use PV = nRT
Solve for P in atm.
To find the pressure exerted on the TA's stomach, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
First, we need to calculate the number of moles of CO2 gas produced from the 8.9 g of dry ice. The molar mass of CO2 is approximately 44.01 g/mol.
Step 1: Calculate the number of moles
Number of moles = mass / molar mass
Number of moles = 8.9 g / 44.01 g/mol
Using a calculator, we find the number of moles to be approximately 0.202 moles.
Next, we convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 311 K
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
Step 2: Rearrange the equation to solve for pressure (P):
P = (nRT) / V
P = (0.202 moles * 0.0821 L*atm/(mol*K) * 311 K) / 1.5 L
Step 3: Calculate the pressure.
Using a calculator, we find the pressure to be approximately 4.09 atm.
Therefore, the pressure exerted on the TA's stomach is approximately 4.09 atm.
To find the pressure exerted on the stomach, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
To solve for pressure, we need to find the number of moles (n). We can calculate the number of moles using the molar mass of CO2 and the given mass (8.9 g) of dry ice.
The molar mass of CO2 is approximately 44 g/mol. We can use this information to determine the number of moles:
n = mass / molar mass
n = 8.9 g / 44 g/mol
n ≈ 0.202 mol
Next, we substitute the values into the ideal gas law equation:
PV = nRT
P * 1.5 L = 0.202 mol * (0.0821 L * atm / K * mol) * 311 K
Simplifying the equation:
P = (0.202 mol * 0.0821 L * atm / K * mol * 311 K) / 1.5 L
P ≈ 2.75 atm
Therefore, the pressure exerted on the TA's stomach is approximately 2.75 atm.