Change in enthalpy when 11.2 dm3 of helium at ntp is heated in a cylinder to 100 degrees celsius

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

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