Hi I have 2 questions I hope someone can help me :)
1)A pocket of gas is discovered in a deep drilling operation. The gas has a temperature of 480c and is at a pressure of 12.8 atm. What volume of gas is required to provide 18.0 L of gas at the surface where the conditions are 22 C and 1.00 atm?
2)What is the volume of 45.0g of nitrogen monoxide, NO, at 20 C and a pressure of 740 mm Hg?
1) 480 C = 753 K; 22 C = 299 K
Let V2 be the volume at the surface (18.0 l) and V1 be the value at the low depth. You want to solve for V1.
P1 V1/T1 = P2 V2/T2, with temperatures in Kelvin.
(12.8 x V1)/753 = (1.00 x 19.0)/ 299
Solve for V1.
2) The molar mass is 16 + 12 = 28.00, so
45.0 g is n = 1.61 moles. That number of moles would occupy
22.4 x 1.61 = 36.06 liters at 1.00 atm and 273 K. At 20 C and 740 mm Hg(which corresponds to 0.974 atm), the volume has to be multiplied by (299/273)(1.000/0.973).
OR you could just use V = n R T/p, with R = 8.206*10^-2 l-atm/(mole K)
is this correct for number 1?
480c = 753k, 22c = 295k
p1 x v1/T1 = p2 x v2/T2
12.8 x v1/753 = 1.00 x 18.0/295
.016 = .061
.016 is V1
No, the calculation for number 1 is not correct. Let me explain the correct steps to solve the problem:
1) Convert the temperatures to Kelvin:
480°C = 480 + 273 = 753 K
22°C = 22 + 273 = 295 K
2) Use the combined gas law equation, which is derived from Boyle's law, Charles's law, and Gay-Lussac's law. The equation is as follows:
(P1 x V1) / T1 = (P2 x V2) / T2
3) Plug the given values into the equation:
(12.8 atm x V1) / 753 K = (1.00 atm x 18.0 L) / 295 K
4) Solve for V1:
(12.8 atm x V1) / 753 K = 0.060676 atm x L / K
Cross multiplying, we get:
12.8 atm x V1 = 753 K x 0.060676 atm x L
V1 = (753 K x 0.060676 atm x L) / 12.8 atm
V1 = 3.542142 L
So, the volume of gas required at the low depth is approximately 3.542 L.
I apologize for the confusion in my initial response.