"Only 2 molecules" of ATP <span>produced during the citric acid cycle
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Answer:
P₂ = 0.09 atm
Explanation:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Given data:
Initial volume = 0.225 L
Initial pressure = 338 mmHg (338/760 =0.445 atm)
Initial temperature = 72 °C (72 +273 = 345 K)
Final temperature = -15°C (-15+273 = 258 K)
Final volume = 1.50 L
Final pressure = ?
Solution:
P₁V₁/T₁ = P₂V₂/T₂
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 0.445 atm × 0.225 L × 258 K / 345 K × 1.50 L
P₂ = 25.83 atm .L. K / 293 K . L
P₂ = 0.09 atm
Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.

Temperature = 
As molar mass of
is 32 g/mol. Hence, the number of moles of
are calculated as follows.

Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.