5 ml of water at 4 degree Celsius (density 1 g/ml) is converted to steam at 110 degree Celsius. What volume will the steam occupy when pressure is 380 Torr?
The volume of the steam can be calculated using the Ideal Gas Law:
V = nRT/P
where n is the number of moles of water, R is the ideal gas constant (0.0821 L·atm/mol·K), T is the temperature in Kelvin (283.15 K), and P is the pressure in atmospheres (0.49 atm).
Therefore, the volume of the steam is:
V = (5 ml * 1 g/ml / 18 g/mol) * 0.0821 L·atm/mol·K * 283.15 K / 0.49 atm
V = 11.7 L
To find the volume of steam at a given pressure, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in Torr)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's calculate the number of moles of water:
Since the density of water is 1 g/ml and we have 5 ml of water, the mass of water is:
mass = density * volume = 1 g/ml * 5 ml = 5 g
The number of moles can be calculated using the molar mass of water (H2O):
molar mass of H2O = 2 * atomic mass of hydrogen + atomic mass of oxygen = 2 * 1 g/mol + 16 g/mol = 18 g/mol
moles of water = mass / molar mass = 5 g / 18 g/mol ≈ 0.278 moles
Next, let's convert the temperature from Celsius to Kelvin:
The conversion from Celsius to Kelvin is done by adding 273.15.
Initial temperature (4 °C) = 4 °C + 273.15 = 277.15 K
Final temperature (110 °C) = 110 °C + 273.15 = 383.15 K
Now we can plug the values into the ideal gas law equation:
Initial pressure (P1) = ?
Initial volume (V1) = 5 ml = 5 cm³ = 5 × 10⁻³ L
Initial moles of water (n1) = 0.278 moles
Initial temperature (T1) = 277.15 K
Final pressure (P2) = 380 Torr
Final volume (V2) = ?
Final moles of water (n2) = 0.278 moles
Final temperature (T2) = 383.15 K
Using the ideal gas law equation, we can find the final volume (V2):
(P1 * V1) / (n1 * T1) = (P2 * V2) / (n2 * T2)
Now we can solve for V2:
V2 = (P1 * V1 * n2 * T2) / (P2 * n1 * T1)
Plugging in the values:
V2 = (1 atm * 5 × 10⁻³ L * 0.278 * 383.15 K) / (380 Torr * 0.278 * 277.15 K)
Converting units:
1 atm = 760 Torr
V2 = (760 Torr * 5 × 10⁻³ L * 0.278 * 383.15 K) / (380 Torr * 0.278 * 277.15 K)
Simplifying:
V2 = (760 * 5 × 10⁻³ * 383.15) / (380 * 277.15)
Calculating:
V2 ≈ 0.03916 L
Therefore, the steam will occupy approximately 0.03916 liters when the pressure is 380 Torr.
To find the volume of steam, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, let's convert the initial temperature of 4 degrees Celsius to Kelvin:
T_initial = 4 degrees Celsius + 273.15 = 277.15 K
Next, let's determine the number of moles of water:
Given the density of water is 1 g/ml, we have:
Mass = Density x Volume = 1 g/ml x 5 ml = 5 g
Since the molar mass of water is approximately 18 g/mol, we can calculate the number of moles(n) of water:
n = Mass / Molar mass = 5 g / 18 g/mol = 0.27778 mol
Now, let's convert the final temperature of 110 degrees Celsius to Kelvin:
T_final = 110 degrees Celsius + 273.15 = 383.15 K
Next, let's convert the pressure from Torr to atm:
1 atm = 760 Torr
Pressure in atm = 380 Torr / 760 Torr/atm = 0.5 atm
Now, we can rearrange the ideal gas law equation to solve for the volume (V):
V = (nRT) / P
Substituting the known values into the equation:
V = (0.27778 mol x 0.0821 L atm/mol K x 383.15 K) / 0.5 atm
Calculating:
V = 17.3 L
Therefore, the volume of steam at 380 Torr will be approximately 17.3 liters.