A given mass of gas occupies a volume of 449 mL at 25°C and 740 mmHg. What will be the new volume at STP?
Use the combined gas law:
P2V2/T2= P1V1/T1
V2=P1V1T2/P2T1
Using the gas equation,
V1=499ml; T1=25C+273K=298
P1=740mmHg
At s.t.p
P2=760mmHg; T2=273K; V2=?
V2= 499*740*273 /760*298
=445.1 ans
To find the new volume of the gas at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation. The ideal gas law equation is:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin
First, let's convert the given temperature from Celsius to Kelvin. To convert Celsius to Kelvin, you add 273.15.
Temperature in Kelvin = 25°C + 273.15 = 298.15 K
Now we need to convert the pressure from mmHg to atm because the value of the ideal gas constant (R) will be consistent with atm unit.
Pressure in atm = 740 mmHg / 760 mmHg (1 atm/760 mmHg) ≈ 0.974 atm
At STP, the temperature is 273.15 K (0°C) and the pressure is 1 atm. So we can rewrite the ideal gas law equation for STP as:
P₁V₁/T₁ = P₂V₂/T₂
Where:
P₁, V₁, and T₁ are the initial conditions
P₂, V₂, and T₂ are the final conditions
Now, let's rearrange the equation to solve for the new volume (V₂):
V₂ = (P₁V₁T₂)/(P₂T₁)
Substituting the values into the equation:
V₂ = (0.974 atm)(449 mL)(273.15 K) / (1 atm)(298.15 K)
Calculate the value:
V₂ ≈ 398.53 mL
Therefore, the new volume of the gas at STP is approximately 398.53 mL.