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