0.33 atm.
By the Ideal Gas Law:
[tex]P \cdot V = n \cdot R\cdot T[/tex]
Where
[tex]T_1 =55.0 \; \textdegree\text{C} = (55.0 + 273.15) = 328.15 \; \text{K}[/tex].
[tex]T_2 = 0 \; \textdegree\text{C} = 273.15 \; \text{K}[/tex] under STP standard conditions.
[tex]P_1 = \dfrac{n \cdot R\cdot T_1}{V}[/tex],
[tex]P_2 = \dfrac{n \cdot R\cdot T_2}{V}[/tex].
The value of [tex]n[/tex], [tex]R[/tex], and [tex]V[/tex] shall be the same in the two equations.
Divide the second equation with the first:
[tex]\dfrac{P_2}{P_1} = \dfrac{n\cdot R\cdot \dfrac{1}{V}}{n\cdot R\cdot \dfrac{1}{V}} \cdot \dfrac{T_2}{T_1} = \dfrac{T_2}{T_1}[/tex].
[tex]P_2 = P_1 \cdot \dfrac{T_2}{T_1} = 0.40 \times \dfrac{273.15}{328.15} = 0.33 \; \text{atm}[/tex].
Answer:
Gay-Lussac's Law; p₁/T₁ = p₂/T₂; 0.33 atm
Step-by-step explanation:
The volume and number of moles are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.
p₁/T₁ = p₂/T₂ Invert each side of the equation
T₁/p₁ = T₂/p₂ Multiply each side by p₂
T₂ = T₁ × p₁/p₂
p₂ = p₁ × T₂/T₁
Data:
p₁ = 0.40 atm; T₁ = 55.0 °C
p₂ = ?; T₂ = 0.0 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = 55.0 + 273.15 = 328.15 K
T₂ = 0.0 + 273.15 = 273.15 K
(b) Calculate the new pressure
p₂ = 0.40 × 273.15/328.15
= 0.40 × 0.832
= 0.33 atm
The pressure drops to 0.33 atm.
