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

Answer:

691.29 K or 418.14 °C

Explanation:

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

$n_{O_2}=1.75 g * \frac{1mol O_2}{32 g O_2} =0.0547 mol$

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

$PV=nRT\\T=\frac{PV}{nR}\\T=\frac{1atm * 3.10 L}{0.0547 mol*0.082 \frac{atm*L}{mol*K} }\\T=691.29 K$

Best regards.