1. the volume of an idea gas is 2.06L at standard temperature and pressure. Calculate the volume at 1.75 atm and 27CELCIUS.
2. Ethane, C2h6 has a molar heat of vaporization of 15kj/mole. How many kilojoules or energy are required to vaporize 5 g of ethane?
Ethane, C2H6, has a molar heat of vaporization of 15 kJ/mole. How many kilojoules or
energy are required to vaporize 5 g of ethane?
To solve these problems, we need to use the ideal gas law and the equation for calculating energy.
1. To find the volume of an ideal gas at a different pressure and temperature, we can use the ideal gas law, which states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/K·mol)
T = temperature in Kelvin
We can solve for the new volume using the equation:
V2 = (P1 * V1 * T2) / (P2 * T1)
Given:
P1 = standard pressure = 1 atm
V1 = 2.06 L
P2 = 1.75 atm
T1 = standard temperature = 273.15 K (0 degrees Celsius + 273.15)
T2 = 27 degrees Celsius + 273.15 = 300.15 K
Plugging in these values, we can calculate the new volume:
V2 = (1 atm * 2.06 L * 300.15 K) / (1.75 atm * 273.15 K)
V2 ≈ 2.27 L
Therefore, the volume of the ideal gas at 1.75 atm and 27 degrees Celsius is approximately 2.27 L.
2. To calculate the energy required to vaporize a substance, we can use the equation:
Energy = molar heat of vaporization * moles of the substance
Given:
Molar heat of vaporization of ethane = 15 kJ/mol
Mass of ethane = 5 g
Molar mass of ethane (C2H6) = 30 g/mol
To find the moles of ethane, we can use the equation:
Moles = Mass / Molar mass
Moles = 5 g / 30 g/mol
Moles ≈ 0.167 mol
Now, we can calculate the energy required to vaporize 5 g of ethane:
Energy = 15 kJ/mol * 0.167 mol
Energy ≈ 2.505 kJ
Therefore, approximately 2.505 kJ of energy are required to vaporize 5 g of ethane.