If you have one mole of a gas at 3.00 atm in a 3.0 L container what is the temperature of the gas?

n = 3.0 moles

V = 60.0 L

T = 400 K

From PV = nRT, you can find P

P = nRT/V = (3.0 mol)(0.0821 L-atm/K-mol)(400 K)/60.0L

P = 1.642 atm = 1.6 atm (to 2 significant figures)

Explanation:

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

the answer would be c