Determine the volume of 1.00 mole of a gas under the following conditions: t = 273 k, p = 2.00 atm: v = l t = 546 k, p = 1.00 atm: v = l t = 409.5 k, p = 1.50 atm: v = l

T = 273 K, P = 2.00 atm: V = 11.2 L  
T = 546 K, P = 1.00 atm: V = 44.8 L  
T = 409.5 K, P = 1.50 atm: V = 22.4 L  
The ideal gas law is:
 PV = nRT
 where
 P = pressure
 V = volume
 n = number of moles
 R = ideal gas constant
 T = absolute temperature
 
 There's a lot of different values for R depending upon the units involved.
Given that we're using liters and atm, the constant I'll use is 0.082057338
L*atm/(K*mol)

 
 Let's solve for volume
 PV = nRT
 V = nRT/P 
 Let's substitute the known values for n and R.
 V = nRT/P
 V = 1.00 mol * 0.082057338 L*atm/(K*mol) * T/P
 V = 0.082057338 L*atm/K * T/P 
 Now simply substitute the given values and calculate. I'll do the full calculation for the 1st problem, and then simply give the answers for the remaining 2 problems. 
 T = 273 K, P = 2.00 atm: V = 11.2 L 
 V = 0.082057338 L*atm/K * T/P
 V = 0.082057338 L*atm/K * 273 K/2.00 atm
 V = 22.40165327 L*atm /2.00 atm
 V = 11.20082664 L
 Rounding to 3 significant figures gives 11.2 L. 
 T = 546 K, P = 1.00 atm: V = 44.8 L 
 T = 409.5 K, P = 1.50 atm: V = 22.4 L

The answer is d.potassium

257 grams h2o i think