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PLEASE HELP(100 points!) A sample of N2 gas was placed in a flexible 12.0L container at 32 degrees C at a pressure of 2.3 atm. The container was compressed to a volume of 4.0L and heat was added until the temperature reached 49 degrees C. What is the new pressure inside the container? Show your work.

Respuesta :

kayv22
To find the new pressure inside the container, we can use the combined gas law equation:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

Let's plug in the values we have:

P1 = 2.3 atm (initial pressure)
V1 = 12.0 L (initial volume)
T1 = 32°C + 273.15 = 305.15 K (initial temperature)

V2 = 4.0 L (final volume)
T2 = 49°C + 273.15 = 322.15 K (final temperature)

Now, we can solve for P2, the new pressure inside the container. Let me do the calculations for you:

(2.3 atm * 12.0 L) / (305.15 K) = (P2 * 4.0 L) / (322.15 K)

Now, let's solve for P2:

P2 = (2.3 atm * 12.0 L * 322.15 K) / (305.15 K * 4.0 L)

P2 ≈ 7.68 atm

So, the new pressure inside the container is approximately 7.68 atm.
semsee45

Answer:

7.3 atm

Explanation:

To find the new pressure inside the container, we can use the combined gas law.

Combined Gas Law

[tex]\boxed{\sf \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}}[/tex]

where:

  • P₁ is the initial pressure.
  • V₁ is the initial volume.
  • T₁ is the initial temperature (in kelvin).
  • P₂ is the final pressure.
  • V₂ is the final volume.
  • T₂ is the final temperature (in kelvin).

Given values:

  • P₁ = 2.3 atm
  • V₁ = 12.0 L
  • T₁ = 32°C
  • V₂ = 4.0 L
  • T₂ = 49°C

Since the temperatures have been provided in Celsius, it is necessary to convert them to kelvin by adding 273.15:

[tex]\sf T_1 = 32 + 273.15=305.15[/tex]

[tex]\sf T_2 = 49 + 273.15=322.15[/tex]

Rearrange the formula to isolate P₂:

[tex]\sf P_2=\dfrac{P_1V_1T_2}{T_1V_2}[/tex]

Now, substitute the values into the formula and solve for P₂:

[tex]\sf P_2=\dfrac{2.3\cdot 12.0 \cdot 322.15}{305.15 \cdot 4.0}\\\\\\P_2=\dfrac{8891.34}{1220.6}\\\\\\P_2=7.2844011142...\\\\\\P_2=7.3\; atm\;(2\;s.f.)[/tex]

Therefore, the new pressure inside the container is 7.3 atm (2 s.f.)

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