1 g of water on evaporation at atmospheric pressure forms 1671 cm3 of steam. Heat of vaporisation at this pressure is 540 cal g-1 . The increase in internal energy is

Question

1 g of water on evaporation at atmospheric pressure forms 1671 cm3 of steam. Heat of vaporisation at this pressure is 540 cal g-1 . The increase in internal energy is (A) 250 cal (B) 500 cal (C) 1000 cal (D) 1500 cal

Tardigrade Admin · Accepted Answer

1 g of water ⇒ 1 cc of water Volume of liquid =VL=1 c c=10-6 m 3 Volume of vapours =VV=1671 cc =1671 × 10-6 m 3 ⇒ Δ V=VV-VL=1670 × 10-6 m 3 ⇒ W=p Δ V=105(1670 × 10-6)=167 J ⇒ W=(167/4.18) cal =40 cal Further, Q=m L ⇒ Q=(1 g)(540 cal g-1)=540 cal According to first law of thermodynamics Q=Δ U+W ⇒ 540=Δ U+40 ⇒ Δ U=500 cal

Q. 1g of water on evaporation at atmospheric pressure forms 1671cm3 of steam. Heat of vaporisation at this pressure is 540calg−1 . The increase in internal energy is

250 cal

500 cal

1000 cal

1500 cal

Q. $1 g$ of water on evaporation at atmospheric pressure forms $1671 c m^{3}$ of steam. Heat of vaporisation at this pressure is $540 c a l g^{- 1}$ . The increase in internal energy is