a gas has a volume of 410 mL at 27 degrees c and 740 mm Hg. what volume would this gas occupy at STP?
According to ideal gas equation [(p1*v1)\t1]=[(p2*v2)/t2]
thus,
[(740*410)/300]=[(760*V)/273]
so,by solving the above eqn. we get
V=363.28 ml
To find the volume of a gas at STP (Standard Temperature and Pressure), we need to use the ideal gas law, which states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
At STP, the values are defined as:
P = 1 atmosphere (atm)
T = 0 degrees Celsius (°C) or 273.15 Kelvin (K)
R = 0.0821 liter * atmosphere / (mole * Kelvin)
First, let's convert the given temperature of 27 degrees Celsius to Kelvin:
T(°C) = 27°C
T(K) = T(°C) + 273.15 = 27 + 273.15 = 300.15 K
Now, we can set up the equation using the given values:
P1 = 740 mm Hg, which need to be converted to atm:
P1(atm) = P1(mm Hg) / 760 mm Hg/atm = 740 / 760 = 0.973684 atm
V1 = 410 mL, which needs to be converted to liters:
V1(L) = V1(mL) / 1000 mL/L = 410 / 1000 = 0.41 L
T1 = 300.15 K
P1V1 / T1 = P2V2 / T2
Since we want to find V2 (the volume at STP), we can rearrange the equation:
V2 = (P1V1T2) / (T1P2)
Now, let's substitute the values into the equation:
P2 = 1 atm
T2 = 273.15 K
V2 = (0.973684 * 0.41 * 273.15) / (300.15 * 1)
V2 = 0.3663 L
Therefore, the gas would occupy a volume of approximately 0.3663 liters at STP.