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 find the pressure at 145 deg C, we can use the combined gas law equation:
(P1)(V1) / (T1) = (P2)(V2) / (T2)
Since n and V do not change, we can cancel them out:
(P1) / (T1) = (P2) / (T2)
Plugging in the given values:
(P1) / (100 + 273) K = (P2) / (145 + 273) K
Now we can solve for P2:
P2 = (P1) * (145 + 273) K / (100 + 273) K
P2 = (1.00 atm) * (145 + 273) K / (100 + 273) K
P2 ≈ 1.977 atm
Therefore, the pressure in atmospheres when the temperature of the steam reaches 145 deg C is approximately 1.977 atm.
To determine the pressure at which the steam in the autoclave will reach 145°C, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atmospheres)
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 convert the given temperatures to Kelvin:
100°C = 100 + 273.15 = 373.15 K
145°C = 145 + 273.15 = 418.15 K
Since the number of moles (n) and the volume (V) do not change, we can write the equation as:
P1V1 / T1 = P2V2 / T2
Substituting the given values:
P1 (1.00 atm) * V1 / T1 (373.15 K) = P2 * V1 / T2 (418.15 K)
Since V1/V1 will cancel out, we have:
P1 / T1 = P2 / T2
Rearranging the equation to solve for P2 (pressure at 145°C):
P2 = (P1 * T2) / T1
Plugging in the given values:
P2 = (1.00 atm * 418.15 K) / 373.15 K
Calculating this expression and rounding to three significant figures, the pressure (P2) is:
P2 = 1.122 atm
So, the pressure in the autoclave will reach 1.12 atm when the temperature of the steam is 145°C.
To determine the pressure at which the temperature of the steam in the autoclave reaches 145°C, we can use the Ideal Gas Law. The Ideal Gas Law equation is:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Given that n and V do not change, we can rearrange the equation as follows:
P1/T1 = P2/T2
Let's plug in the given values:
P1 = 1.00 atm (initial pressure)
T1 = 100°C (initial temperature)
T2 = 145°C (final temperature)
Converting temperatures to Kelvin:
T1 = 100°C + 273.15 = 373.15 K
T2 = 145°C + 273.15 = 418.15 K
Now we can solve for P2:
P1/T1 = P2/T2
P2 = (P1 × T2) / T1
P2 = (1.00 atm × 418.15 K) / 373.15 K
Calculating the value:
P2 = 1.121 atm
Therefore, the pressure at which the temperature of the steam in the autoclave reaches 145°C is approximately 1.121 atm.