18.1-1. diffusion of methane through helium. a gas of ch4 and he is contained in a tube at 101.32 kpa pressure and 298 k. at one point, the partial pressure of methane is pa1 = 60.79 kpa, and at a point 0.02 m distance away, pa2 = 20.26 kpa. if the total pressure is constant throughout the tube, calculate the flux of ch4 (methane) at steady state for equimolar counterdiffusion.

5.521 × 10⁻² mol/m².s

Explanation:

Given:

Pressure of the Methane and Helium gas = 101.32 kPa

Temperature of the Methane and Helium gas = 298 K

Partial pressure of Methane,  pA₁ = 60.79 kPa

Partial pressure of Methane at point 0.02 m away,  pA₂  = 20.26 kPa

Now,

Molar flux is given as:

Here,

= 0.675 × 10⁻⁴ m²/s (for He-CH4 at 101.32 kPa and 298 K)

Z₂ - Z₁ = 0.02 m

R is the ideal gas constant = 8.314 J/mol.K

T is the temperature =  298 K

On substituting the respective values, we get

or

= 5.521 × 10⁻² mol/m².s

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

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