The gas inside of a container exerts #4 Pa# of pressure and is at a temperature of #220 ^o C#. If the temperature of the gas changes to #150 ^oC# with no change in the container's volume, what is the new pressure of the gas?

Answer 1

#P_f=3.43" "Pa#

#P_i:"initial pressure (Pa)"# #P_f:"final pressure (Pa)"# #T_i:"initial temperature (Kelvin)"# #T_f:"final temperature (Kelvin)"#

#"given data:"# #P_i=4Pa# #P_f=?# #T_i=220+273=493 " "^o K# #T_f=150+273=423" "^o K#

#"We can use the Gay-Lussac's Law to calculate the problem"#

#P_i/T_i=P_f/T_f#

#4/493=P_f/423#

#P_f=(4*423)/493#

#P_f=3.43" "Pa#

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Answer 2

Lily Adams

You can use the Gay-Lussac's law equation:

(P_1 / T_1 = P_2 / T_2)

Where: (P_1 =) initial pressure (4 Pa), (T_1 =) initial temperature (220 °C), (P_2 =) new pressure (unknown), (T_2 =) new temperature (150 °C).

Rearrange the equation and plug in the values:

(P_2 = P_1 \times (T_2 / T_1))

(P_2 = 4 , \text{Pa} \times (150 / 220))

(P_2 ≈ 2.7273 , \text{Pa})

Therefore, the new pressure of the gas is approximately 2.7273 Pa.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.