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What is the temperature of a sample of 0.00925 mol of oxygen gas at 0.865 atm in a 296 ml container?

Options:
A. 356 K
B. 366 K
C. 376 K
D. 386 K

1 Answer

7 votes

Final answer:

The temperature of a sample of oxygen gas at 0.865 atm in a 296 ml container can be determined using the ideal gas law PV = nRT, where we convert the volume to liters, plug in the values for pressure, volume, number of moles, and the ideal gas constant, and solve for T (temperature).

Step-by-step explanation:

To determine the temperature of a sample of oxygen gas under given conditions, we can use the ideal gas law, which is PV = nRT. Here, P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvins (K).

First, we need to convert the volume from milliliters to liters since the R constant is typically given in liters. So, 296 mL is converted to 0.296 L. Now we can rearrange the ideal gas law to solve for T (temperature), resulting in T = PV/(nR).

We are given:

  • P (pressure) = 0.865 atm
  • V (volume) = 0.296 L
  • n (number of moles) = 0.00925 mol
  • R (ideal gas constant) = 0.0821 L·atm/(mol·K)

Using these values:

T = (0.865 atm × 0.296 L) / (0.00925 mol × 0.0821 L·atm/(mol·K))

Calculating this gives us the temperature T. Let's find out which of the options matches the calculated T value.

User Arthur Rubens
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