An ideal gas has a molecular mass of 10 grams per mole, and a specific internal
energy of 300 kilojoules per kilogram when its temperature is 373 kelvin. What is the specific enthalpy of this gas?
Before we go to calculate the specific enthalpy, let’s consider what enthalpy is in
and of itself. This term which is sometimes represented by the letter capital 𝐻 refers to the total
heat of a system. Mathematically, this total heat of a system — its enthalpy — is equal to its internal
energy plus its pressure times its volume.
To get a picture of what enthalpy means, imagine a system coming into existence out
of thin air. In order to create that system, we need some amount of energy; that’s Δ𝑈. And then in order for this system to have a spatial extent — means to push back
against the pressure and volume of the atmosphere it’s in — that too takes
energy. And that energy added to the internal energy is the system’s enthalpy. That’s enthalpy.
But what about specific enthalpy? Specific enthalpy is simply equal the total enthalpy divided by the mass of the
system. It’s this enthalpy we want to solve for for the ideal gas we’re considering. Based on these two equations, we can say that the specific enthalpy we want to solve
for is equal to Δ𝑈, our system’s internal energy, plus its pressure times its
volume all over its mass.
We can rewrite the right-hand side of this expression as two terms. And we realize that in our problem statement, we’re given the specific internal
energy of our system — that’s Δ𝑈, the internal energy — over the system’s mass
𝑚. This value we’re told is 300 kilojoules per kilogram.
So now, we’ll focus on the 𝑃𝑣 over 𝑚 term for our specific enthalpy. Recalling that we’re working with an ideal gas, we realize that we can re-express 𝑃
times 𝑣 in terms of the ideal gas law. This law says that for an ideal gas 𝑃𝑣 is replaceable with a number of moles in our
gas times the gas constant times the gas’s temperature.
If we consider this term 𝑛, the number of moles in our system, we know that that’s
equal to the total mass of our system divided by its molar mass. And we’re told that this ideal gas has a molar mass of 10 grams per mole. With this substitution in for the number of moles 𝑛 and knowing the temperature of
our system 𝑇, we’re about ready to plug in and solve for specific enthalpy.
Before we do, let’s look up the value of the gas constant 𝑅, which when we do, we
find it’s equal to 8.314 joules per mole kelvin. Plugging all these values in and replacing the mass 𝑚 of our system with an
expression of 𝑥 kilograms since we don’t know just how much the mass is, but we
know its units, we’ll see that lots of the units in this expression begin to cancel
The units of kelvin cancel out as do the units of moles and the units of kilogram in
the left-hand side of our numerator. Also the total mass of our system the number 𝑥 cancels out in numerator and
denominator. We’re left in the end with a value in units of joules per kilogram, which is what we
expect for a specific enthalpy.
So the units remaining are joules per kilogram and the numbers remaining are 8.314
times 373 over 0.010. To two significant figures, this equals 310 times 10 to the third joules per
We then add this result to the specific internal energy of our system, 300 kilojoules
per kilogram, to get our final answer of 610 kilojoules per kilogram. That’s the specific enthalpy of this gas.