1 g of water (volume 1 cm 3 ) becomes 1671 cm 3 of steam when boiled at a pressure of 1 atm. The latent heat of vaporization is 540 cal / g, then the external work done is: (1 atm =1.013 × 105 N / m 2)

Question

1 g of water (volume 1 cm 3 ) becomes 1671 cm 3 of steam when boiled at a pressure of 1 atm. The latent heat of vaporization is 540 cal / g, then the external work done is: (1 atm =1.013 × 105 N / m 2) (A) 499. 7 J (B) 40.3 J (C) 169.2 J (D) 128.57

Tardigrade Admin · Accepted Answer

Work done, W = p Δ V = 1.013 × 105 × (1671 - 1) = 1.013 × 105 × 1670 × 10-6 = 169.2 J

Q. $1 g$ of water (volume $1 c m^{3}$ ) becomes $1671 c m^{3}$ of steam when boiled at a pressure of $1 a t m$ . The latent heat of vaporization is $540 c a l / g$ , then the external work done is:
$(1 a t m = 1.013 \times 1 0^{5} N / m^{2})$

499. 7 J

40.3 J

169.2 J

128.57

Work done, $W = p Δ V$
$= 1.013 \times 1 0^{5} \times (1671 - 1)$
$= 1.013 \times 1 0^{5} \times 1670 \times 1 0^{- 6}$
$= 169.2 J$

Q. 1g of water (volume 1cm3 ) becomes 1671cm3 of steam when boiled at a pressure of 1atm. The latent heat of vaporization is 540cal/g, then the external work done is: (1atm=1.013×105N/m2)