What volume in liters does 0.750 g of N2 occupy at 19.5 oC and 748 torr?
Assuming the gas is ideal, we can use the Ideal Gas Law:
PV = nRT
where
P = pressure in atm
V = volume in liters
n = number of moles
R = universal gas constant = 0.0821 L-atm/mol-K
T = temperature in Kelvin
Useful units conversion:
1 atm = 760 torr
Kelvin = Celsius + 273
The molar mass of N2 = 28 g/mol. To get n, we divide the given mass (0.750 g) by the molar mass of N2 (the 28 g/mol).
Then, convert the given to their appropriate units and substitute them to the ideal gas equation to solve for the volume.
hope this helps :3
To solve this problem, 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 in Kelvin.
In this case, we are given the mass of nitrogen gas (N2), the temperature in Celsius, and the pressure. We need to convert the temperature to Kelvin and find the number of moles of N2 in order to solve for the volume.
1. Convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 19.5 + 273.15
T(K) = 292.65 K
2. Convert the mass of N2 to moles using its molar mass:
Molar mass of N2 = 28 g/mol
Moles (n) = mass (g) / molar mass (g/mol)
n = 0.750 g / 28 g/mol
n ≈ 0.02679 mol
3. Convert the pressure from torr to atmosphere (atm):
1 atm = 760 torr
P(atm) = P(torr) / 760
P(atm) = 748 torr / 760
P(atm) ≈ 0.98316 atm
4. Now we can rearrange the ideal gas law equation to solve for V (volume):
V = (nRT) / P
V = (0.02679 mol) * (0.0831 L·atm/(mol·K)) * (292.65 K) / (0.98316 atm)
V ≈ 0.694 L
So, 0.750 g of N2 occupies approximately 0.694 liters at 19.5 °C and 748 torr.