2.2 g of nitrous oxide (N2O) gas is cooled at a constant pressure of 1

1.	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
Answer» 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 \(\triangle U = -72.49 KJ\)

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

Answer»

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 \(\triangle U = -72.49 KJ\)