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The gas inside of a container exerts #18 Pa# of pressure and is at a temperature of #70 ^o K#. If the pressure in the container changes to #48 Pa# with no change in the container's volume, what is the new temperature of the gas?

Answer 1

Olivia Askew

The new trmperature is #=186.7K#

We apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#

#P_1=18Pa#

#T_1=70K#

#P_2=48Pa#

#T_2=P_2/P_1*T_1#

#=48/18*70=186.7K#

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Answer 2

Nora Arzate

The new temperature of the gas is approximately 186 degrees Kelvin.

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Answer 3

Axel Acevedo

To find the new temperature (T₂) of the gas, we can use the ideal gas law:

[P_1V_1 / T_1 = P_2V_2 / T_2]

Where: (P_1 = 18 Pa) (initial pressure) (T_1 = 70 K) (initial temperature) (P_2 = 48 Pa) (final pressure) (V_1 = V_2) (no change in volume)

Rearranging the equation to solve for (T_2):

[T_2 = (P_2 * T_1) / P_1]

Substituting the values:

[T_2 = (48 * 70) / 18]

[T_2 = 186.67 K]

Therefore, the new temperature of the gas is (186.67) Kelvin.

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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.