Final answer:
The change in enthalpy when heating 11.2 dm³ of helium at NTP to 100 degrees Celsius is -896 J.
Step-by-step explanation:
The change in enthalpy when heating 11.2 dm³ of helium at NTP (normal temperature and pressure) to 100 degrees Celsius can be calculated using the equation:
ΔH = q = mcΔT
Where ΔH is the change in enthalpy, q is the heat energy absorbed, m is the mass of helium, c is the specific heat capacity of helium, and ΔT is the change in temperature.
Since helium is a monoatomic ideal gas, its specific heat capacity at constant pressure (Cp) is approximately 5/2 R, where R is the ideal gas constant. Therefore, the equation becomes:
ΔH = q = (5/2)RnΔT
Given that 11.2 dm³ of helium is equivalent to 0.0112 m³ and using the ideal gas equation PV = nRT with standard pressure (1 atm) and temperature (273 K), we can calculate the number of moles (n) of helium:
n = PV/RT = (1 atm)(0.0112 m³)/(0.0821 L·atm/(mol·K))(273 K) = 0.523 mol
Substituting the values into the equation, we have:
ΔH = q = (5/2)(8.314 J/(mol·K))(0.523 mol)(100 - 273) K)
Calculating ΔH, we get:
ΔH = q = -896 J