Calculate the volume of one mole of gas at STP(Standard Temperature & Pressure). Given: P=101 325 Pa
n=1mol
R=8,31 J K-1 mol-1
T=273K
Given: P1=785mm Hg
P2=760mm Hg
T1=293 K
T2=273 K
V1=60cm3
Nope. You have a unit problem. If you use the value of R in J/K*mol
then pressure has to be in Pa and volume to be in m^3
V=nRT/P=1*8.31*273/101.2 E3
V=22.39 E-3 m^3 or 22.4dm^3
To calculate the volume of one mole of gas at STP (Standard Temperature and Pressure), we can use the Ideal Gas Law equation: PV = nRT.
For the first scenario, where P = 101,325 Pa, n = 1 mole, R = 8.31 J K-1 mol-1, and T = 273 K:
1. Convert the pressure from Pascals to atmospheres (atm):
P = 101,325 Pa / 101,325 = 1 atm
2. Plug the values into the Ideal Gas Law equation and solve for V:
V = (nRT) / P
V = (1 mol * 8.31 J K-1 mol-1 * 273 K) / 1 atm
V ≈ 22.4 L
So, the volume of one mole of gas at STP is approximately 22.4 liters.
For the second scenario, we can use Boyle's Law, which states that P1V1 = P2V2, to find the final volume (V2) when the pressure and temperature change:
1. Convert the pressures from mm Hg to atm:
P1 = 785 mm Hg / 760 mm Hg/atm = 1.0329 atm
P2 = 760 mm Hg / 760 mm Hg/atm = 1 atm
2. Plug the values into Boyle's Law equation and solve for V2:
P1V1 = P2V2
(1.0329 atm)(60 cm3) = (1 atm)(V2)
V2 ≈ 61.88 cm3
So, the final volume (V2) in the second scenario is approximately 61.88 cm3.