Thermodynamics - Result Question 1
1. $1 g$ of water, of volume $1 cm 3$ at $100^{\circ} C$, is converted into steam at same temperature under normal atmospheric pressure $(» 1 \times 10^{5} Pa)$. The volume of steam formed equals $1671 cm^{3}$. If the specific latent heat of vaporisation of water is $2256 J / g$, the change in internal energy is,
[NEET Odisha 2019]
(a) $2256 J$
(b) $2423 J$
(c) $2089 J$
(d) $167 J$
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Solution:
- (c) $\Delta Q=2256 \times 1=2256 J$
$\Delta W=P[V _{\text{steam }}-V _{\text{water }}] \quad[\because d w=p d v]$
$=1 \times 105[1671-1] \times 106=167 J$
As $\Delta Q=\Delta U+\Delta W$
$2256=\Delta U+167$
$\Delta U=2089 J$
