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Sagot :
Given that N and V are constants:
[tex]P=\frac{NRT}{V}\\(400)=(8.31)(110)\frac{N}{V}\\400=914.59\frac{N}{V}\\\frac{400}{914.59}=\frac{N}{V}\\\\P=(8.31)(235)\frac{N}{V}\\P=(1953.90)\frac{400}{914.59}\\P=854.55\\854.55=about~854.46[/tex]
C. 854.46 kPa
[tex]P=\frac{NRT}{V}\\(400)=(8.31)(110)\frac{N}{V}\\400=914.59\frac{N}{V}\\\frac{400}{914.59}=\frac{N}{V}\\\\P=(8.31)(235)\frac{N}{V}\\P=(1953.90)\frac{400}{914.59}\\P=854.55\\854.55=about~854.46[/tex]
C. 854.46 kPa
Answer : (C) 854.46 KPa.
Solution : Given,
Initial pressure = 400 KPa
Initial temperature = 110 K
Final temperature = 235 K
According to the Gay-Lussac's law, the absolute pressure is directly proportional to the absolute temperature at constant volume of an ideal gas.
P ∝ T
Formula used :
[tex]\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}[/tex]
where,
[tex]P_{1}[/tex] = initial pressure
[tex]P_{2}[/tex] = final pressure
[tex]T_{1}[/tex] = initial temperature
[tex]T_{2}[/tex] = final temperature
Now put all the values in above formula, we get
[tex]\frac{400}{P_{2}}=\frac{110}{235}[/tex]
By rearranging the terms, we get the value of new/final pressure.
[tex]P_{2}[/tex] = 854.5454 KPa [tex]\approx[/tex] 854.55 KPa
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