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10. A quantity of a gas has an absolute pressure of 400 kPa and an absolute temperature of 110 degrees kelvin. When the temperature of the gas is raised to 235 degrees kelvin, what is the new pressure of the gas? (Assume that there''s no change in volume.)
A. 1.702 kPa
B. 3.636 kPa
C. 854.46 kPa
D. 510 kPa

Sagot :

Since3122015
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

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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