Bacteria and viruses are inactivated by temperatures above 145 deg C in an autoclave. An autoclave contains steam at 1.00 atm and 100 deg C At what pressure, in atmospheres, will the temperature of the steam in the autoclave reach 145 deg C if n and V do not change ? Express the pressure to three significant figures and include the appropriate units.
To solve this problem, we will use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
We are given:
P = 1.00 atm
T1 = 100 °C
T2 = 145 °C
We need to find the new pressure, P2.
First, we need to convert the temperatures to Kelvin:
T1 = 100 °C + 273.15 = 373.15 K
T2 = 145 °C + 273.15 = 418.15 K
Since n and V do not change, we can write the equation as:
P1/T1 = P2/T2
Solving for P2, we have:
P2 = (P1 × T2) / T1
P2 = (1.00 atm × 418.15 K) / 373.15 K
P2 ≈ 1.122 atm
Therefore, the pressure of the steam in the autoclave when the temperature reaches 145 °C is approximately 1.122 atm.
To find the pressure of the steam in the autoclave when its temperature reaches 145°C, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant (0.0821 L•atm/mol•K)
T = temperature in Kelvin
First, let's convert the given temperatures to Kelvin:
100°C + 273.15 = 373.15 K (initial temperature)
145°C + 273.15 = 418.15 K (final temperature)
Since n and V do not change, we can rewrite the equation as:
P1V1/T1 = P2V2/T2
We know:
P1 = 1.00 atm
V1 = V (unchanged)
T1 = 373.15 K
T2 = 418.15 K
Now, rearranging the equation and solving for P2:
P2 = (P1 * T2 * V1) / (T1 * V2)
Since V1 = V2, we can simplify the equation to:
P2 = (P1 * T2) / T1
Plugging in the values:
P2 = (1.00 atm * 418.15 K) / 373.15 K
P2 ≈ 1.12 atm
Therefore, the pressure of the steam in the autoclave will be approximately 1.12 atm when its temperature reaches 145°C.
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure in atm
V is the volume in liters (which remains constant in this case)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
In this problem, we are given:
Initial pressure, P1 = 1.00 atm
Initial temperature, T1 = 100°C = 373 K
Final temperature, T2 = 145°C = 418 K
Since the number of moles (n) and volume (V) do not change, they will cancel out when comparing the two conditions.
Using the equation for both initial and final states, we can write:
P1 * V = n * R * T1
P2 * V = n * R * T2
Dividing the two equations, we get:
P2/P1 = T2/T1
Substituting the given values, we have:
P2/1.0 atm = 418 K/373 K
Cross multiplying, we get:
P2 = 1.0 atm * (418 K/373 K) = 1.123 atm
Rounding to three significant figures, the pressure in the autoclave when the temperature reaches 145°C is approximately 1.12 atm.