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?

Answer 1

New temperature is #462.86^oK#

As there is no change in Volume #V#, #P/T# is always constant, where #p# is Pressure and #T# is Temperature in degree Kelvin.
As initially container exerts #42Pa# of pressure and is at a temperature of #270^oK#, #P/T=42/270#. As pressure in the container changes to #72Pa#, if Temperature changes to #T_1#
#72/T_1=42/270#
Hence #T_1=(72*270)/42=462.86^oK#
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Answer 2

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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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.

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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