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Question

Physics Question on Thermodynamics terms

1g1\, g of water on evaporation at atmospheric pressure forms 1671cm31671\,c{{m}^{3}} of steam. Heat of vaporisation at this pressure is 540calg1540\,cal\,{{g}^{-1}} . The increase in internal energy is

A

250 cal

B

500 cal

C

1000 cal

D

1500 cal

Answer

500 cal

Explanation

Solution

1g1\, g of water
1cc\Rightarrow 1 cc of water Volume of liquid
=VL=1cc=106m3=V_{L}=1 c c=10^{-6} \,m ^{3} Volume of vapours
=VV=1671cc=1671×106m3=V_{V}=1671\, cc =1671 \times 10^{-6} \,m ^{3}
ΔV=VVVL=1670×106m3\Rightarrow \Delta V=V_{V}-V_{L}=1670 \times 10^{-6} \, m ^{3}
W=pΔV=105(1670×106)=167J\Rightarrow W=p \Delta V=10^{5}\left(1670 \times 10^{-6}\right)=167 \,J
W=1674.18cal=40cal\Rightarrow W=\frac{167}{4.18} cal =40 \, cal
Further, Q=mLQ=m L
Q=(1g)(540calg1)=540cal\Rightarrow Q=(1 g)\left(540\, cal\, g^{-1}\right)=540 \, cal
According to first law of thermodynamics
Q=ΔU+WQ=\Delta U+W
540=ΔU+40\Rightarrow 540=\Delta U+40
ΔU=500cal\Rightarrow \Delta U=500 \, cal