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in Chemical thermodynamics by (75 points)
edited by

2.2 g of nitrous oxide (N2O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, ΔU is

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by (48.4k points)

Molecular weight of nitrous oxide (N2O) = 44 g/mol

Number of moles of N2O = \(\frac{2.2}{44} = 0.05 mol\) 

Specific heat capacity of N2O = 880 J/g.K

∴ Heat release on cooling from 310K to 270 K 

q = -2.2 x 880 x 40 K

q = -77.44 KJ   

Negative sign indicates heat released.

∴ Work done by surrounding on system to compares the gas from 217.1 ml to 167.75 ml at constant pressure.

It means work done by surrounding is irreversible.

\(\therefore W = -P \times \triangle V\)

\(= - 1.01 \times 10^5 \times (167.75 - 217.1 ml)\)

\(= 1.01 \times 10^5 \times 0.049 J\)

\(W = 4.95 kJ\)

Using 1st Law of thermodynamics

\(\triangle U = q + W\)

\(\triangle U = -77 .44 KJ + 4.95KJ\)

\(\triangle U = -72.49 KJ\)

Hence change in internal energy \(\triangle U = -72.49 KJ\)

by (75 points)
Molar specific heat is given to be 100 kJ/mol the q was getting long so I cut it out my answer is -150.65 but it should be -195 idk where am I going wrong the q is 49.35 and W=-200 in my case
by (42.1k points)
Given, molar heat capacity = 100KJ/ mol.K
Number of moles of N2O = 2.2/44 = 0.05 mol.
∴ q = nCm△T
Cm = molar heat capacity
n = Number of moles
△T = change in temperature
q = 0.05 x 100 x (270 - 310) x 10^3
q = -200KJ
Work done by the surrounding on system
W = -PDV
 = - 1.013 x 10^5 x (167. 75 - 217.1)
= 4.99 KJ
∴ △U = q + W
= -200 KJ + 4.99 KJ
△U ≃ -195 KJ

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