A gas is heated from 263 K to 298 K, and the volume is increased from 24 L to 35 L. If the original pressure is 1 atm, what would the final pressure be? *

Answers

Answer 1
Answer:

Since for closed system moles of the gas is always conserved

so as per idea gas equation we can say

n = (PV)/(RT)

so here we can say

n_1 = n_2

so here we have

(P_1V_1)/(RT_1) = (P_2V_2)/(RT_2)

as we know that

T_1 = 263 K

T_2 = 298 K

V_1 = 24 L

V_2 = 35 L

P_1 = 1 atm

now from above equation

(1*24)/(263) = (P*35)/(298)

on solving above equation we have

P = 0.78 atm

so here pressure will be 0.78 atm

Answer 2
Answer:

Final answer:

Based on the combined form of the ideal gas law, the final pressure of the gas, given the changes in volume and temperature, would be approximately 0.725 atm.

Explanation:

The question concerns the change in conditions of a gas and asks you to determine the final pressure. This deals with the combined form of the ideal gas law, which states that the product of the initial pressure and volume, divided by the initial temperature, equals the product of the final pressure and volume, divided by the final temperature (P₁V₁/T₁ = P₂V₂/T₂).

Given that the initial pressure P₁ is 1 atm, initial volume V₁ is 24 L, initial temperature T₁ is 263 K, final volume V₂ is 35 L, and final temperature T₂ is 298 K, we can substitute these values into the equation to solve for the final pressure P₂.

Therefore, P₂ = P₁V₁T₂ / V₂T₁ = (1 atm × 24 L × 298 K) / (35 L × 263 K) ≈ 0.725 atm. So, the final pressure of the gas would be approximately 0.725 atm.

Learn more about Ideal Gas Law here:

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Answers

Answer:

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Answers

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Answers

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Answers

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Answers

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Answers

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Hope it helps!