The gas inside of a container exerts #18 Pa# of pressure and is at a temperature of #360 ^o K#. If the pressure in the container changes to #24 Pa# with no change in the container's volume, what is the new temperature of the gas?
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Using the ideal gas law equation ( PV = nRT ), where:
- ( P ) is the pressure,
- ( V ) is the volume,
- ( n ) is the number of moles of gas,
- ( R ) is the ideal gas constant, and
- ( T ) is the temperature in Kelvin.
Since the volume remains constant, the equation can be simplified to ( P_1/T_1 = P_2/T_2 ), where ( P_1 ) and ( T_1 ) are the initial pressure and temperature, and ( P_2 ) and ( T_2 ) are the final pressure and temperature, respectively.
Given: ( P_1 = 18 , \text{Pa} ), ( T_1 = 360 , \text{K} ), ( P_2 = 24 , \text{Pa} ).
Substituting these values into the equation:
[ \frac{18}{360} = \frac{24}{T_2} ]
[ T_2 = \frac{24 \times 360}{18} ]
[ T_2 = 480 , \text{K} ]
So, the new temperature of the gas is ( 480 , \text{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.
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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