If 39.5 mol of an ideal gas is at 4.05 atm at 47.20 ∘C, what is the volume of the gas?

Answer 1

Use the ideal gas law:

PV = nRT,

where P is the pressure, V is the volume,T is the temperature (in Kelvin), n is the moles of gas, and R is the gas constant. Units are very important. Values of P, V, and T vary wildly, but the ones used in the equation must match the units of the gas constant, R. A value of R may be chosen among many possibilities. Each option has the same intrinsic value, but differ in the units.

In this problem, we are given units of C, moles, and atm. The temperature must be converted to Kelvin (add 273.15 to C). Since we are using atm for pressure, find a value of R with atm in the unit. I'll use R = 0.08206 L*atm*/(K*mol).

Rearrange:

PV = nRT

V = nRT/P

V = (39.5 moles)*(0.08206 L*atm*/(K*mol))*(320.4K)/(4.05 atm)

Cancel the units:

V = (39.5 moles)*(0.08206 L*atm*/(K*mol))*(320.4K)/(4.05 atm)

V = (39.5)*(0.08206 L)*(320.4)/(4.05)

We are left with only L, liters. This is what we want, so do the math:

V = 256 liters

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A Perspective:

At STP, a gas occupies 22.4 liters/mole. This isn't at STP, but we can use this information to help judge the validity of the value we obtained.

At STP, (39.5 moles gas)*(22.4 L/mole) = 884.8 liters (STP). This seems like a big difference over the 256 liters calculated above, but note that the pressure has more than quadrupled compared to standard pressure of 1 atm. This would reduce the volume by the same factor, by four times. (884.8 L/4) = 221.2 liters. That is getting a lot closer to the 256 L we obtained. The difference is due to the higher temperature of 47.2C compared to standard temperature of 0C. The volume of 256 L may be adjusted for this temperature deviation from STP:

(221.2 L)*(320.3K/273.2K) = 259 liters. This is close (enough, for me) to the 256 liters in the original calculation to give confidence we took the correct steps.