First slide

Change in internal energy

Question

One gram of water of volume 1 c.c becomes 1671 c.c of steam when boiled at a pressure of one atmosphere. The latent heat of vaporization of water is 540 cal/gm. Find the increase in internal energy (1 at. pressure = 10⁵ N/m²)

Moderate

A

540 Cal
B

500.24 Cal
C

167 Cal
D

39.66 Cal

Solution

$ΔU = Q - W$
Here $W = P \times (V_{2} - V_{1})$
$= 10^{5} \times (1671 - 1) \times 10^{- 8} J = 167 J$
and Q = 1x540cal
$∴ Δ U = (540 - \frac{167}{4 .2}) cal = 500 .24 cal .$

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