A container with a volume of #12 L# contains a gas with a temperature of #150^o C#. If the temperature of the gas changes to #620 ^o K# without any change in pressure, what must the container's new volume be?

Answer 1

new volume #~~17.6 L#

Because the pressure don't change we apply #V_1/T_1=V_2/T_2# where V (L), T(K) Hence #12/423=V_2/620# #V_2=12*620/423# #V_2~~17.6 L#

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Answer 2

Adrian Ayala

To solve this problem, we'll use the combined gas law equation:

(V_1 / T_1 = V_2 / T_2)

Given: (V_1 = 12 L) (T_1 = 150°C = 423 K) (T_2 = 620 K)

Rearranging the equation to solve for (V_2): (V_2 = (V_1 * T_2) / T_1)

Substitute the given values: (V_2 = (12 L * 620 K) / 423 K)

(V_2 ≈ 17.62 L)

So, the container's new volume must be approximately 17.62 liters.

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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.