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The gas inside of a container exerts #15 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

Amelia Adams

The new pressure is #13.2 Pa#

Apply the gas law.

#P_1/T_1=P_2/T_2#

#P_1=15 Pa#

#T_1=220#ºC#=493K#

#T_2=150#ºC#=423K#

#P_2=15/493*434=13.2 Pa#

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Answer 2

Sebastian Acosta

You can use the ideal gas law, (PV = nRT), where (P) is pressure, (V) is volume, (n) is the number of moles of gas, (R) is the ideal gas constant, and (T) is temperature in Kelvin.

Since the volume remains constant, (V) does not change.

(P_1 / T_1 = P_2 / T_2)

Where: (P_1) = initial pressure, (T_1) = initial temperature in Kelvin, (P_2) = final pressure, and (T_2) = final temperature in Kelvin.

Converting temperatures to Kelvin:

(T_{\text{initial}} = 220 + 273 = 493 , K) (T_{\text{final}} = 150 + 273 = 423 , K)

Substitute the values into the equation:

(15 / 493 = P_2 / 423)

Solve for (P_2):

(P_2 = (15 * 423) / 493)

(P_2 ≈ 12.88 , 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.