1.Calculate the volume occupied by 3.2g of oxygen at 300K and 100KNm^-2.(R=8.3J/mol/K,Oxygen =16).
2.Calculate the volume occupied by 0.125mol of Oxygen at 27°C and a pressure of 2.02×10^5Nm^2.(1mole of gas occupies 22.4dm^3 at Standard temperature and pressure,where standard temperature is =1.01×10^5Nm^-2)
1. 100 kN/m^2 = 0.998 atm, R = 0.08206 L*atm/mol*K , n = 3.2/32
PV = nRT
0.998*V = (3.2/32)*0.08206 L*atm/mol*K * 300 K
Solve for V in liters
For 2. You surely don't mean standard temperature is =1.01×10^5Nm^-2)
Can someone solve number 2 question please
To calculate the volume occupied by a gas, 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
Let's solve each problem step-by-step.
1. Calculate the volume occupied by 3.2g of oxygen at 300K and 100KNm^-2.
First, we need to convert the mass of oxygen from grams to moles using its molar mass. The molar mass of oxygen (O2) is 32 g/mol.
Given:
Mass of oxygen = 3.2g
Molar mass of oxygen = 32 g/mol
Number of moles (n) = Mass / Molar mass
n = 3.2g / 32 g/mol
n = 0.1 mol
Now, let's substitute the values into the ideal gas law equation:
PV = nRT
V = nRT / P
V = (0.1 mol)(8.3 J/mol/K)(300K) / (100000 N/m^2)
V = 24.9 J / 100000 N/m^2
V = 2.49 x 10^-4 m^3
Therefore, the volume occupied by 3.2g of oxygen at 300K and 100KNm^-2 is approximately 2.49 x 10^-4 m^3.
2. Calculate the volume occupied by 0.125mol of Oxygen at 27°C and pressure of 2.02×10^5 Nm^2.
First, we need to convert the temperature from degrees Celsius to Kelvin by adding 273.15.
Given:
Number of moles (n) = 0.125 mol
Temperature (T) = 27°C + 273.15 = 300.15K
Pressure (P) = 2.02 x 10^5 N/m^2
Now, let's substitute the values into the ideal gas law equation:
PV = nRT
V = nRT / P
V = (0.125 mol)(8.3 J/mol/K)(300.15K) / (2.02 x 10^5 N/m^2)
V = 24.68 J / 2.02 x 10^5 N/m^2
V = 1.22 x 10^-4 m^3
Therefore, the volume occupied by 0.125mol of Oxygen at 27°C and a pressure of 2.02×10^5 Nm^2 is approximately 1.22 x 10^-4 m^3.
To calculate the volume occupied by a gas, you 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 the gas, R is the ideal gas constant, and T is the temperature.
1. To calculate the volume occupied by 3.2g of oxygen at 300K and 100KNm^-2, you first need to convert the mass of the oxygen to moles.
First, calculate the number of moles of oxygen:
Number of moles (n) = mass (m) / molar mass (M)
Molar mass of oxygen (O2) = 16g/mol (given)
n = 3.2g / 16g/mol = 0.2 moles
Next, plug the values into the ideal gas law equation and solve for volume (V):
PV = nRT
P = 100KN/m^2 = 100,000 N/m^2 (given)
R = 8.3 J/mol/K (given)
T = 300K (given)
n = 0.2 moles (calculated)
V = (nRT) / P
V = (0.2 moles * 8.3 J/mol/K * 300K) / 100,000 N/m^2
Now, you can calculate the volume:
V = 0.498 dm^3
Therefore, the volume occupied by 3.2g of oxygen at 300K and 100KNm^-2 is approximately 0.498 dm^3.
2. To calculate the volume occupied by 0.125 moles of oxygen at 27°C and a pressure of 2.02×10^5 Nm^2, you can use the same method.
First, convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T = 27°C + 273.15 = 300.15K
Next, plug the values into the ideal gas law equation and solve for volume (V):
PV = nRT
P = 2.02×10^5 N/m^2 (given)
R = 8.3 J/mol/K (given)
T = 300.15K (calculated)
n = 0.125 moles (given)
V = (nRT) / P
V = (0.125 moles * 8.3 J/mol/K * 300.15K) / 2.02×10^5 N/m^2
Now, you can calculate the volume:
V = 0.0197 m^3
Since 1m^3 = 1000 dm^3, you can convert the volume to dm^3:
V = 19.7 dm^3
Therefore, the volume occupied by 0.125 moles of oxygen at 27°C and a pressure of 2.02×10^5 Nm^2 is approximately 19.7 dm^3.