Water is brought to boil under a pressure of 1.0 atm. When an electric current of 0.50 A from a 12 V supply is passed for 300 s through a resistance in thermal contact with it, it is found that 0.798 g of water is vaporised. Calculate the molar internal energy change at boiling point (373.15 K).

Question

Water is brought to boil under a pressure of 1.0 atm. When an electric current of 0.50 A from a 12 V supply is passed for 300 s through a resistance in thermal contact with it, it is found that 0.798 g of water is vaporised. Calculate the molar internal energy change at boiling point (373.15 K). (A)  37.5 kJ mol-1  (B)  3.75 kJ mol-1  (C)  42.6 kJ mol-1  (D)  4.26 kJ mol-1

Tardigrade Admin · Accepted Answer

Δ H=work done =i× V× t =0.50× 12× 300=1800 J=1.8 kJ Molar enthalpy of vaporisation, Δ Hm=(Δ H/moles of H2O)=(Δ H/nH2O) =(1.8 kJ/(0.798)18)=40.6 kJ mol-1 Δ Hm=Δ Em+pΔ V Δ Hm=Δ Em+Δ ngRT Δ Hm=Δ Em+RT [∴ Δ ng=1for H2O(l)H2O(g)] ∴ Molar internal energy change, Δ Em=Δ Hm-RT Δ Em=40.6-8.314× 10-3× 373.15 =37.5 kJ mol-1

Q. Water is brought to boil under a pressure of 1.0 atm. When an electric current of 0.50 A from a 12 V supply is passed for 300 s through a resistance in thermal contact with it, it is found that 0.798 g of water is vaporised. Calculate the molar internal energy change at boiling point (373.15 K).

$37.5 k J m o l^{- 1}$

$3.75 k J m o l^{- 1}$

$42.6 k J m o l^{- 1}$

$4.26 k J m o l^{- 1}$

Q. Water is brought to boil under a pressure of 1.0 atm. When an electric current of 0.50 A from a 12 V supply is passed for 300 s through a resistance in thermal contact with it, it is found that 0.798 g of water is vaporised. Calculate the molar internal energy change at boiling point (373.15 K).

37.5kJmol−1

3.75kJmol−1

42.6kJmol−1

4.26kJmol−1

$37.5 k J m o l^{- 1}$

$3.75 k J m o l^{- 1}$

$42.6 k J m o l^{- 1}$

$4.26 k J m o l^{- 1}$