A sample of N2 gas occupies a volume 180 mL at STP. What volume, in L, will it occupy at 640 mm Hg and 295 K?
(P1V1/T1)= (P2V2/T2)
T must be in kelvin.
To find the volume of N2 gas at 640 mm Hg and 295 K, we can use the combined gas law equation:
P1 * V1 / T1 = P2 * V2 / T2
Where:
P1 = pressure at initial state (STP = 1 atm)
V1 = volume at initial state (180 mL)
T1 = temperature at initial state (STP = 273 K)
P2 = pressure at final state (640 mm Hg)
V2 = volume at final state (unknown)
T2 = temperature at final state (295 K)
First, we need to convert the given values to consistent units. We'll use the following conversions:
1 atm = 760 mm Hg
1 L = 1000 mL
Converting the initial state conditions:
P1 = 1 atm
V1 = 180 mL = 0.180 L
T1 = 273 K
Converting the final state conditions:
P2 = 640 mm Hg / 760 mm Hg/atm = 0.8421 atm
T2 = 295 K
Substituting these values into the equation:
(1 atm) * (0.180 L) / (273 K) = (0.8421 atm) * V2 / (295 K)
Simplifying the equation:
0.180 / 1 = (0.8421 / 1) * V2 / (295 / 273)
0.180 = (0.8421) * V2 / (1.08)
Cross-multiplying:
0.180 * 1.08 = 0.8421 * V2
0.1944 = 0.8421 * V2
Dividing both sides by 0.8421:
V2 = 0.1944 / 0.8421
V2 ≈ 0.2307 L
Therefore, the volume of N2 gas at 640 mm Hg and 295 K is approximately 0.2307 L.
To solve this problem, we need to use the combined gas law equation:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
Where:
P₁ and P₂ are the initial and final pressures (in atm)
V₁ and V₂ are the initial and final volumes (in L)
T₁ and T₂ are the initial and final temperatures (in Kelvin)
First, we need to convert the given volume from milliliters (mL) to liters (L):
180 mL = 180/1000 L = 0.18 L
Next, we need to convert the given pressure from mm Hg to atm:
1 atm = 760 mm Hg
640 mm Hg = 640/760 atm = 0.8421 atm (rounded to 4 decimal places)
Now, we can substitute the values into the combined gas law equation and solve for V₂:
(1 * 0.18) / (273) = (0.8421 * V₂) / (295)
Cross multiplying the equation gives us:
0.18 * 295 = 0.8421 * V₂
53.1 = 0.8421 * V₂
Dividing both sides of the equation by 0.8421 gives us:
V₂ ≈ 63.0736 L (rounded to 4 decimal places)
Therefore, the sample of N2 gas will occupy approximately 63.0736 liters at 640 mm Hg and 295 K.