5. What volume would 27g of neon at 2 atm pressure and 50°C occupy?
I have solved the problem correctly, but I am not sure if I have the correct units on this conversion so that everything will cancel out. Also, is it right to have 20g Ne/1mole K? I seem to be missing something with the 20g Ne.
V = (27g Ne/20g Ne/1mole K)(0.08206 L atm)x(323K)/2 atm = 17.8L
My calculation shows 17.89 L which I would round to 17.9L.
For mol = n you should have
27g Ne x (1 mol Ne/20g Ne) = 13.5 mol Ne. Showing the factor as I've shown it makes more sense than the way you do although yours is correct except for the K. (1 mol K should be 1 mol. The K is not needed.)
Then if you plug in P in atm, R in (L*atm/mol*K) and T in K, V comes out in L.
can you explain how to do t his i am confused
To determine the volume of a gas, 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, R is the ideal gas constant, and T is the temperature in Kelvin.
In this case, we are given the following information:
Pressure (P) = 2 atm
Mass of neon (m) = 27g
Temperature (T) = 50°C
To start, we need to convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 50°C + 273.15
T(K) = 323.15K
Next, we need to calculate the number of moles of neon. To do this, we will use the molar mass of neon, which is approximately 20g/mol:
n = m/M
n = 27g / 20g/mol
n = 1.35 mol
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
V = (nRT) / P
V = (1.35 mol)(0.08206 L·atm/mol·K)(323.15K) / 2 atm
Calculating this expression, we get:
V ≈ 17.67 L
Thus, the volume occupied by 27g of neon at 2 atm pressure and 50°C is approximately 17.67 liters.
Regarding your question about unit conversions, it appears that you might have made a mistake in your calculation. The molar mass of neon is approximately 20g/mol, not 20g Ne/1mol K. By using the correct conversion and substituting all the values with consistent units, you should arrive at the correct answer.