A mass of gas has a volume of 380 cm3 at a pressure of 560 mm Hg and a temperature of 7C. What is its bolume at s.t.p.?
P = 560/760= 0.73684 atm
T = 280 K
At STP (To = 273 and Po = 1 atm), the volume at (P,T) gets mutiplied by
(To/T)(P/Po) = (273/280(1/0.73684)
= 1.323
That makes the new (STP) volume 503 cm^3
273 cm3
ijwdlnrwlker
To find the volume of a gas at standard temperature and pressure (STP), we need to use the ideal gas law equation, which is:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal gas constant
T = Temperature
First, we need to convert the given temperature from Celsius to Kelvin, as the ideal gas law requires the temperature to be in Kelvin.
To do this, we use the formula:
T(K) = T(C) + 273.15
Given temperature, T(C) = 7°C
Converting Celsius to Kelvin: T(K) = 7 + 273.15 = 280.15 K
Now, we have:
P = 560 mm Hg
V = 380 cm^3
T = 280.15 K
Next, we need to convert the pressure and volume to their respective SI units:
1 mm Hg = 1 torr = 133.322 Pa
1 cm^3 = 1 mL
Converting the pressure to pascals (Pa):
P(Pa) = 560 mm Hg * 133.322 Pa/mm Hg
P(Pa) ≈ 74694.32 Pa
Converting the volume to liters (L):
V(L) = 380 cm^3 / 1000 cm^3/L
V(L) = 0.38 L
Substituting the values into the ideal gas law equation:
PV = nRT
Using STP conditions (STP temperature = 273.15 K, STP pressure = 1 atm, R = 0.0821 L*atm/(mol*K))
1 atm = 101325 Pa
We can rearrange the equation to solve for the volume at STP:
V(STP) = (n * R * T(STP)) / P(STP)
Plugging in the values:
V(STP) = (n * 0.0821 L*atm/(mol*K) * 273.15K) / 101325 Pa
Since the number of moles (n) is not given, we cannot determine the exact volume at STP without this information.