The gas inside of a container exerts #48 Pa# of pressure and is at a temperature of #320 ^o K#. If the pressure in the container changes to #60 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, we can determine the new temperature of the gas:
[ P_1 \times V_1 = P_2 \times V_2 ]
Since the volume remains constant:
[ P_1 = \frac{nRT_1}{V} ]
[ P_2 = \frac{nRT_2}{V} ]
Since the volume is constant:
[ \frac{P_1}{T_1} = \frac{P_2}{T_2} ]
Rearranging the equation:
[ T_2 = \frac{P_2 \times T_1}{P_1} ]
Substitute the given values:
[ T_2 = \frac{60 , \text{Pa} \times 320 , \text{K}}{48 , \text{Pa}} ]
[ T_2 = 400 , \text{K} ]
Therefore, the new temperature of the gas is 400 Kelvin.
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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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