# The gas inside of a container exerts #27 Pa# of pressure and is at a temperature of #150 ^o K#. If the pressure in the container changes to #40 Pa# with no change in the container's volume, what is the new temperature of the gas?

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The new temperature is

Apply Gay Lussac's law

The final temperature is

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You can use the ideal gas law, ( PV = nRT ), where:

- ( P ) is the pressure of the gas (in pascals),
- ( V ) is the volume of the gas (in cubic meters),
- ( n ) is the number of moles of the gas,
- ( R ) is the ideal gas constant (8.31 J/(mol·K)),
- ( T ) is the temperature of the gas (in kelvin).

Since the volume remains constant, you can use the relationship ( P_1 / T_1 = P_2 / T_2 ) to solve for the new temperature ( T_2 ).

Given: ( P_1 = 27 ) Pa (initial pressure), ( T_1 = 150 ) K (initial temperature), ( P_2 = 40 ) Pa (final pressure).

Using the equation ( P_1 / T_1 = P_2 / T_2 ), we can solve for ( T_2 ): [ T_2 = \frac{P_2 \times T_1}{P_1} ]

Plugging in the values: [ T_2 = \frac{40 \times 150}{27} ]

[ T_2 ≈ 222.22 , K ]

Therefore, the new temperature of the gas is approximately ( 222.22 , K ).

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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.

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.

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.

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