Given, mass of water, m=lg Volume of 1 g of water =1 cm3=10−6 m3 Volume of 1 g of steam =1671 cm3=1671×10−6 m3 Pressure, p=1×105 Pa Latent heat of vaporization of water, L=2256 J/g Change in volume, ΔV=(1671−1)×10−6 m3 =1670×10−6 m3 Heat supplied, ΔQ=mL=1×2256=2256 J As the steam expands, so the work done in expansion is ΔW=pΔV=1×105×1670×10−6 [from Eq. (i)] =167 J….(iii) According to first law of thermodynamics, ΔQ=ΔU+ΔW⇒ΔU=ΔQ−ΔW=2256−167=2089 J [from Eq. (ii) and (iii)]
1 g of water, of volume 1 ~cm^3 at 100° C is converted into steam at same temperature under normal atmospheric pressure = (simeq 1 × 10^5 ~Pa ). The…
NEET UG 2019 — Physics Thermodynamics
2019mcqhard
1 g of water, of volume 1 cm3 at 100∘C is converted into steam at same temperature under normal atmospheric pressure =(≃1×105 Pa). The volume of steam formed equals 1671 cm3. If the specific latent heat of vaporisation of water is 2256 J/g, the change in internal energy is
Official previous-year question
Held on 30 Apr 2019 · Verified 9 Jul 2026.
Options
- A
2423 J
- B
2089 J
- C
167 J
- D
2256 J
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