IF THE QUANTITY OF OXYGEN OCCUPYING 2.76L CONTAINER AT A PRESSURE OF 0.825 ATM AND 300K IS REDUCED BY ONE-HALF, WHAT IS THE PRESSURE EXERTED BY THE REMAINING GAS

• A. 1.650atm
• B. 0.825atm
• C. 0.413atm ✓
• D. 0.275atm

 

The answer to the question is: C. 0.413atm

To solve this problem, we can use the ideal gas law equation, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

First, let’s calculate the initial number of moles of oxygen in the 2.76L container at a pressure of 0.825 atm and 300K using the ideal gas law:

P1V1 = nRT

(0.825 atm) × (2.76 L) = n × (0.0821 L.atm/mol.K) × (300 K)

n = (0.825 atm) × (2.76 L) / (0.0821 L.atm/mol.K) × (300 K)

n ≈ 0.099 moles

Now, if the quantity of oxygen is reduced by one-half, the new number of moles (n’) will be half of the initial number of moles:

n’ = 0.5 n

n’ = 0.5 × 0.099

n’ ≈ 0.0495 moles

Next, we can use the ideal gas law to find the new pressure (P’) exerted by the remaining gas in the same 2.76L container at 300K:

P’V = n’RT

P’ = (n’RT) / V

P’ = (0.0495 moles × 0.0821 L.atm/mol.K × 300 K) / 2.76 L

P’ ≈ 0.413 atm

Therefore, the pressure exerted by the remaining gas is approximately 0.413 atm.