Answer:
Enthalpy change = 333.3 J
Explanation:
Given data:
Volume of helium = 11.2 dm³ or 11.2 L
Temperature = 100 °C (100+273= 373 K)
Enthalpy change = ?
Solution:
Normal temperature of room = 20°C (20+273 = 293 K)
one mole of gas occupy volume = 22.4 L
ΔT = final temperature - initial temperature
ΔT = 373K - 293K
ΔT = 80K
Number of moles of gas = 11.2 L/22.4 l
Number of moles of gas = 0.5 mol
Enthalpy change = nRΔT
Enthalpy change = 0.5 mol × 8.3j/mol.k ×80 K
Enthalpy change = 333.3 J
