Asample consisting of 2 moles he is expanded isothermally at 0 degrees from 5.0dm3 to 20.0dm3. calculate w, q and deltau for each of the following situations: (i) a reversible expansion of the sample. (ii) an irreversible expansion of the sample against a constant external pressure equal to the final pressure of the gas. (iii) a free expansion (against zero external pressure i. e. in a vacuum) of the sample.

i)

ii)

ii)

Explanation:

As the initial and final states of the sample are the same, the ΔU of the sample is, for the three cases

since

i)Reversibly so can be calculated by  

and because of the first law of thermodynamics

ii)Irreversibly with

we can calculate by the law of ideal gases

then w can be calculated by

and  

iii)a free expansion so (there's no work at vaccum) and

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

im not exactly sure but i think its b!