What pressure is required to reduce 34 mL of a gas at standard conditions to 16 mL at a temperature of 20◦C
P1V1=nRT1
P2V2=nRT2
P2=P1*V1/V2*T2/t1
P2=1atm*34/16*293/273
What pressure is required to reduce 34 mL of
a gas at standard conditions to 16 mL at a
temperature of 20◦C?
0.000026566
To determine the pressure required to reduce the volume of a gas, we can use Boyle's Law. Boyle's Law states that the pressure and volume of a gas are inversely proportional when temperature is kept constant. Mathematically, it can be represented as:
P1 * V1 = P2 * V2
where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
In this case, we are given:
Initial volume (V1) = 34 mL
Final volume (V2) = 16 mL
Temperature (T) = 20°C
Since we are working at standard conditions, we can assume the initial and final pressures are equal to the standard pressure of 1 atmosphere (atm).
Let's substitute the values we have into Boyle's Law equation:
(1 atm) * (34 mL) = (P2) * (16 mL)
Simplifying the equation:
34 mL = (P2) * 16 mL
To find the value of P2, we can rearrange the equation:
P2 = (34 mL) / (16 mL)
P2 ≈ 2.125
Therefore, the pressure required to reduce 34 mL of gas at standard conditions to 16 mL at a temperature of 20°C is approximately 2.125 atm.