A sample of helium behaves as an ideal gas as it is heated at constant pressure from 278 K to 353 K. If 40 J of work is done by the gas during this process, what is the mass of the helium sample? The universal gas constant is 8.31451 J/mol · K.

Answer in units of g.

To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (constant)
V = volume (unknown in this case)
n = number of moles of the gas
R = universal gas constant
T = temperature (in Kelvin)

We are given:
P = constant (as the gas is heated at constant pressure)
R = 8.31451 J/mol · K
T1 = 278 K
T2 = 353 K
Work done by the gas = 40 J

Since the pressure is constant, we can rewrite the equation as:

V1/n1T1 = V2/n2T2

We know that the volume of an ideal gas is directly proportional to the number of moles. Therefore, we can rewrite the equation as:

V1/T1 = V2/T2

Now, let's solve for the volume ratio:

V1/V2 = T1/T2

Plugging in the given values:
V1/V2 = 278 K / 353 K

V1/V2 ≈ 0.788

Since the work done by the gas during the process is given, we can calculate the work done using the equation:

Work = P * (V2 - V1)

Since the pressure is constant, the equation simplifies to:

Work = P * V2 - P * V1

40 J = P * V2 - P * V1

Since V1/V2 ≈ 0.788, we can rewrite the equation as:

40 J = P * V2 - P * (0.788 * V2)

Now, we need to find the mass of the helium sample. To do that, we need the number of moles. We can use the ideal gas law equation to find the number of moles:

PV = nRT

Rearranging the equation:

n = PV / RT

Since we have the volume ratio V1/V2 ≈ 0.788 and the initial volume V1, we can find the final volume V2:

V2 = V1 / (V1/V2) = V1 / 0.788

Now, we can substitute the volume values into the equation:

n = P * V2 / RT

Substituting P = constant and V2 = V1 / 0.788:

n = P * (V1 / 0.788) / RT

Now, we can calculate the number of moles.

Next, we can calculate the mass of the helium sample using the molar mass of helium, which is 4.002602 g/mol.

Mass of helium sample = number of moles * molar mass of helium

Finally, we can substitute the number of moles into the equation to find the mass of the helium sample in grams.