Answer: Option D) 6.43 L
Explanation:
Given that,
volume (V) = ?
Pressure (P) = 2.02 atm
Temperature (T) = 15 °C
[Convert 15°C to Kelvin by adding 273
15°C + 273 = 288K]
Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1
Number of moles of N2 (n) = ?
molar mass of N2 (m.m) = 28.01348 g/mole
mass in grams of N2 = 15.4 grams
Recall that Number of moles
= mass in grams / molar mass
n = 15.4 grams / 28.01348 g/mole
n = 0.55 moles
Then, apply ideal gas equation
pV = nRT
2.02 atm x V = 0.55 moles x (0.0821 atm L K-1 mol-1 x 288K)
2.02 atm•V = 13 atm•L
Divide both sides by 2.02 atm
2.02 atm•V/2.02 atm = 13 atm•L/2.02 atm
V = 6.43 L
Thus, the volume of nitrogen gas is 6.43 liters
Answer:
The correct answer is D. 6, 43L.
Explanation:
First we calculate the number of moles in 14 grams of Nitrogen using the molar mass, we convert the unit of temperature in Celsius into Kelvin and then we apply the ideal gas law by solving for volume (using the ideal gas constant R = 0.082 l atm / K mol):
28,01348 g----1 mol N2
15,4g ----x= (15,4g x1 mol N2)/28,01348 g= 0,55 mol N2
0°C= 273K ----> 15°C= 273 + 15= 288 K
PV= nRT ---> V= (nRT)/P
V= (0,55 mol x 0,082 l atm/ K molx 288K)/2,02 atm
V=6, 43 L
