The gas inside of a container exerts #42 Pa# of pressure and is at a temperature of #270 ^o K#. If the pressure in the container changes to #72 Pa# with no change in the container's volume, what is the new temperature of the gas?
New temperature is
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Using the ideal gas law equation (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:
(P_1V = nRT_1) and (P_2V = nRT_2)
Since (V) and (n) remain constant:
(\frac{P_1}{T_1} = \frac{P_2}{T_2})
Rearranging for (T_2):
(T_2 = \frac{P_2 \cdot T_1}{P_1})
Substituting the given values:
(T_2 = \frac{72 , \text{Pa} \times 270 , \text{K}}{42 , \text{Pa}} = 464.29 , \text{K})
So, the new temperature of the gas is approximately (464.29 , \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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