What pressure is required to reduce 75 mL of a gas at standard conditions to 11 mL at a temperature of 27 ◦C?
what pressure would i use?
To determine the pressure required to reduce the volume of a gas from 75 mL to 11 mL at a temperature of 27 ◦C, you can use the combined gas law. The combined gas law relates the initial and final pressure, volume, and temperature of a gas. The formula for the combined gas law is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 is the initial pressure (standard pressure in this case, which is 1 atm)
V1 is the initial volume (75 mL)
T1 is the initial temperature (standard temperature in this case, which is 273 K)
P2 is the final pressure (what you want to find)
V2 is the final volume (11 mL)
T2 is the final temperature (27 ◦C, which needs to be converted to Kelvin)
To solve for P2, plug in the known values into the formula:
(1 atm * 75 mL) / (273 K) = (P2 * 11 mL) / (27 + 273 K)
Now, rearrange the equation to solve for P2:
P2 = ((1 atm * 75 mL) / (273 K)) * ((27 + 273 K) / 11 mL)
Next, perform the calculation to obtain the result for P2. Remember to convert the temperature to Kelvin before performing the calculation.
P2 = (75 mL * 300 K) / (273 K * 11 mL)
Simplifying the equation yields:
P2 ≈ 2.358 atm
Therefore, the pressure required to reduce the gas volume from 75 mL to 11 mL at a temperature of 27 ◦C is approximately 2.358 atm.