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maw [93]
4 years ago
12

A mixture of N2, H2 and He have mole fractions of 0.25, 0.65, and 0.10, respectively. What is the partial pressure of N2 if the

total pressure of the mixture is 3.9 atm?
Physics
2 answers:
Thepotemich [5.8K]4 years ago
6 0

Answer:

.98

Explanation:

matrenka [14]4 years ago
3 0

Answer: The partial pressure of N_2 if the total pressure of the mixture is 3.9 atm is 0.975 atm

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

p_A=x_A\times P_T

where, x = mole fraction of nitrogen in solution = 0.25

p_A = partial pressure  of nitrogen = ?

P_T = Total pressure = 3.9 atm

Putting in the values :

p_A=0.25\times 3.9=0.975atm

The partial pressure of N_2 is 0.975 atm

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Answer:

1.23×10⁸ m

Explanation:

Acceleration due to gravity is:

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When the object is on the surface of the Earth, a = g and r = R.

g = GM / R²

When the object is at height i above the surface, a = 1/410 g and r = i + R.

1/410 g = GM / (i + R)²

Divide the first equation by the second:

g / (1/410 g) = (GM / R²) / (GM / (i + R)²)

410 = (i + R)² / R²

410 R² = (i + R)²

410 R² = i² + 2iR + R²

0 = i² + 2iR − 409R²

Solve with quadratic formula:

i = [ -2R ± √((2R)² − 4(1)(-409R²)) ] / 2(1)

i = [ -2R ± √(1640R²) ] / 2

i = (-2R ± 2R√410) / 2

i = -R ± R√410

i = (-1 ± √410) R

Since i > 0:

i = (-1 + √410) R

R = 6.37×10⁶ m:

i ≈ 1.23×10⁸ m

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