Question

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 ‘–x’ J. The value of ‘x’ is ____. [nearest integer]
(Given : atomic mass of N = 14 g mol^–1 and of O = 16 g mol^–1
Molar heat capacity of N₂O is 100 J K^–1 mol^–1)

Answer 1

Correct Answer: 195

To solve the problem, we need to find the change in internal energy (ΔU) for the given process using the relation: ΔU = q + w, where q is the heat exchanged and w is the work done.

First, calculate the number of moles of N₂O:

1. $\text{Molar mass of N}=14 g mol-1, \text{and O}=16 g mol-1$

2. Molar mass of N₂O = (2 × 14) + 16 = 44 g/mol

3. Number of moles = $\genfrac{}{}{0.1ex}{}{2.2}{44}$ = 0.05 mol

Next, calculate q using the molar heat capacity (C_p) and temperature change (T₂ - T₁):

4. C_p = 100 J K^–1 mol^–1

5. q = n × C_p × (T₂ - T₁)

6. q = 0.05 mol × 100 J K^–1 mol^–1 × (270 K - 310 K)

7. q = 0.05 × 100 × (-40) = -200 J

To find work done (w):

8. w = −PΔV = −P(V₂ - V₁)
Given: V₁ = 217.1 mL = 0.2171 L, V₂ = 167.75 mL = 0.16775 L, P = 1 atm

9. w = −(1 atm) × (0.16775 - 0.2171) L

10. Convert atm-L to J: 1 atm-L = 101.3 J

11. w = −(1 × –0.04935) × 101.3 = 4.999 J

Finally, compute ΔU:

12. ΔU = q + w = −200 J + 5 J = −195 J

The value of ‘x’ is therefore 195, which lies within the expected range (195, 195).