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Each of the following processes caused the gas volume to double, as shown. For each process, tell how the remaining gas variable changed or state that it remained fixed: A. T doubles at fixed P. B. Tand n are fixed C. At fixed T, the reaction is CD2(g) → C(g) + D2 (g) D. At fixed P, the reaction is Az(g) + B2(g) → 2 AB (g)

User Anson
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Answer: (A) V doubles, (B) P decreases by double, (C) n increases by double, (D) T increases by double

Step-by-step explanation:

With the given information, we are looking at a problem that is related to the ideal gas law: PV = nRT, where

  • P = pressure
  • V = volume
  • n = amount of the gas, in moles
  • R = ideal gas constant (value and units vary depending on the units of the other terms)
  • T = temperature

For this problem, we do not need to know any discrete values. Rather, we can use PV = nRT as a guiding equation to predict changes if any of the variables are changed. We are told the gas volume doubles in each of the following processes.

A. If T doubles, while P remains fixed, that means in order for the PV term to remain proportional to T in the ideal gas equation, V has to double since P is fixed. We can assume n and R are not changing in each of the following processes since these properties are inherent to the gas molecules present, leaving the relationship as PV ∝ T

B. If T and n are fixed as V doubles, this means P will decrease by double as well. Since R is a constant, that leaves the following relationship PV = K, where K represents some constant value comprised of the product of T, n, and R. Therefore, P and V are inversely proportional to each other (P ∝ 1/V).

C. At fixed T and given the reaction CD2(g) → C(g) + D2 (g), we see that the number of moles in gas molecules (n) increases via a decomposition reaction. Therefore, n doubles as V doubles.

D. At fixed P and given the reaction Az(g) + B2(g) → 2 AB(g), we see that a combination reaction occurs but the number of moles of the gas stays the same. This is because there is.1 mole of Az and 1 mole of B2 as reactants, compared to 2 moles of AB product. Thus, we started with 2 moles as reactants and ended with 2 moles of product. Since we are told that V doubles, then T increases by double to maintain the ideal gas laws.

User Ptah
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