JEE MAIN - Chemistry (2020 - 2nd September Morning Slot - No. 1)
The internal energy change (in J) When 90 g of
water undergoes complete evaporation at
100oC is ____________.
(Given : $$\Delta $$Hvap for water at 373 K = 41 kJ/mol,
R = 8.314 JK–1 mol–1)
(Given : $$\Delta $$Hvap for water at 373 K = 41 kJ/mol,
R = 8.314 JK–1 mol–1)
Answer
189494TO189495
Explanation
H2O(l) ⇌ H2O(g)
90 gm of H2O = $${{90} \over {18}}$$ moles of H2O = 5 moles of H2O
$$\Delta $$Hvap = $$\Delta $$U + $$\Delta $$ngRT
$$ \Rightarrow $$ $$\Delta $$U = $$\Delta $$Hvap - $$\Delta $$ngRT
= 41000 - 1$$ \times $$8.314$$ \times $$373
= 37898.878
For 5 moles, $$\Delta $$U = 37898.878 $$ \times $$ 5 = 189494.39 Joule
90 gm of H2O = $${{90} \over {18}}$$ moles of H2O = 5 moles of H2O
$$\Delta $$Hvap = $$\Delta $$U + $$\Delta $$ngRT
$$ \Rightarrow $$ $$\Delta $$U = $$\Delta $$Hvap - $$\Delta $$ngRT
= 41000 - 1$$ \times $$8.314$$ \times $$373
= 37898.878
For 5 moles, $$\Delta $$U = 37898.878 $$ \times $$ 5 = 189494.39 Joule
Comments (0)
