Question

What is the temperature (°C) of 1.75 g of O2 gas occupying 3.10 L at 1.00 atm? Enter your answer in the box provided.

Answer 1

691.29 K or 418.14 °C

Step-by-step explanation:

Hello, at first the moles of oxygen gas are required:

`n_(O_2)=1.75 g * (1mol O_2)/(32 g O_2) =0.0547 mol`

Now, based on the ideal gas equation, we solve for the temperature:

`PV=nRT\\T=(PV)/(nR)\\T=(1atm * 3.10 L)/(0.0547 mol*0.082 (atm*L)/(mol*K) )\\T=691.29 K`

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