The gas inside of a container exerts #24 Pa# of pressure and is at a temperature of #90 ^o K#. If the pressure in the container changes to #52 Pa# with no change in the container's volume, what is the new temperature of the gas?

Answer 1

#T_2=195" "^o K#

#P_1/T_1=P_2/T_2#
# P_1=24" Pa, "T_1=90^o K ," P_2=52Pa#
#24/90=52/T_2#
#T_2=(90*52)/24#
#T_2=195" "^o K#
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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, (R) is the gas constant, and (T) is temperature:

(P_1V_1 = P_2V_2) (since volume remains constant)

(T_2 = \frac{P_2V_1}{P_1}\times T_1)

Substituting the given values:

(T_2 = \frac{52 , Pa \times V_1}{24 , Pa} \times 90 , K)

(T_2 = \frac{52}{24} \times 90 , K)

(T_2 = 195 , K)

Therefore, the new temperature of the gas is (195 , 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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