Respuesta :
Answer:
[tex]P_{gauge} = 1.3\times 10^4 \Pa[/tex]
Explanation:
P₁ = 1.01 x 10⁵ Pa
T₁ = 20⁰ C = 20 + 273 = 293 K
T₂ = 58° C = 58 + 273 = 331 K
using gas equation
[tex]\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}[/tex]
[tex]\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}[/tex]
size will not change V₁ = V₂
now,
[tex]\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}[/tex]
[tex]P_2 = \dfrac{P_1}{T_1}\times T_2[/tex]
[tex]P_2 = \dfrac{1.01 \times 10^5}{293}\times 331[/tex]
P₂ = 1.14 x 10⁵ Pa
[tex]P_{gauge} = P_2 - P_{atm}[/tex]
[tex]P_{gauge} = 1.14\times 10^5 - 1.01\times 10^5[/tex]
[tex]P_{gauge} = 1.3\times 10^4 \Pa[/tex]
The gauge pressure inside such a bulb when it is hot; [tex]P_{gauge}=1.3*10^4[/tex]
Given:
P₁ = 1.01 x 10⁵ Pa
T₁ = 20⁰ C = 20 + 273 = 293 K
T₂ = 58° C = 58 + 273 = 331 K
To find:
P₂=?
Gay-Lussac's Law:
It states that for a constant volume, the pressure is directly proportional to absolute temperature: P alpha T; also stated as P/T = K, where K is a constant, and similarly, [tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]
So from using the gas equation & substituting the given values:
[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}\\\\P_2=\frac{P_1}{T_1} *T_2\\\\P_2=\frac{1.01*10^5}{293}*331\\\\P_2=1.14*10^5Pa[/tex]
Thus,
[tex]P_{gauge}=P_2-P_{atm}\\\\P_{gauge}=1.14*10^5-1.01*10^5\\\\P_{gauge}=1.3*10^4[/tex]
Thus, the gauge pressure inside such a bulb when it is hot; [tex]P_{gauge}=1.3*10^4[/tex]
Find more information about "Gay-Lussac's Law" here:
brainly.com/question/24691513
