Two liter glass flask contains some mercury. It is found that at all temperatures the volume of the air inside the flask remains same. The volume of the mercury in side the flask is
$(α f o r g l a s s = 9 \times 10^{- 6} /^{0} C, γ_{H g} = 1.8 \times 10^{- 4} /^{0} C)$ $x \times 10^{- 6} m^{3}$ where x=

Question

Two liter glass flask contains some mercury. It is found that at all temperatures the volume of the air inside the flask remains same. The volume of the mercury in side the flask is

$(α f o r g l a s s = 9 \times 10^{- 6} /^{0} C, γ_{H g} = 1.8 \times 10^{- 4} /^{0} C)$ $x \times 10^{- 6} m^{3}$ where x=

Infinity Learn · Accepted Answer

at all temperatures the volume of the air inside the flask remains same VCγ$$C=VL$$γL ; here V is volume of container and volume of liquid; γ coefficient of volume expansion of container and liquid  $$VL=VC$$γCγL given $$VC=2liter$$;γ$$C=3$$α$$=3x9x10-6/C$$  $$0$$;γ$$L=1.8x10-4/C$$  $$0$$, substitute $$VL=2literx3x9x10-61.8x10-4$$        since $$1liter=10-3m3$$ $$VL=300x10-6m3=volume$$ of mercury in the flask

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Two liter glass flask contains some mercury. It is found that at all temperatures the volume of the air inside the flask remains same. The volume of the mercury in side the flask isα for glass=9×10−6/0C, γHg=1.8×10−4/0Cx×10−6m3 where x=

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