The ideal gas law is
pY = nRT
where
p = pressure
V = volume
T = absolute temperature
n = number of moles
R = gas constant, 8.314 J*(mol-K)
Therefore, the number of moles is
[tex]n= \frac{pV}{RT} [/tex]
The given volume is
V = 1.75 L = 1.75 x 10⁻³ m³
State 1:
T₁ = 25 C = 25+273 = 298 K
p₁ = 225 kPa = 225 x 10³ N/m²
State 2:
T₂ = 10 C = 283 K
p₂ = 185 kPa = 185 x 10³ N/m²
The loss in moles of gas from state 1 to state 2 is
[tex]\Delta n = \frac{V}{R} ( \frac{p_{1}}{T_{1}} - \frac{p_{2}}{T_{2}}) [/tex]
V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N
p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)
p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)
Therefore
Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))
= 0.0213 mol
Answer: b. 0.0213 mol
