The freezing point of a liquid decreases when | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to the Clausius-Clapeyron equation,the change in freezing point with pressure is given by $\frac{dT}{dP} = \frac{T(V_2 - V_1)}{L}$,where

Vedclass · Accepted Answer

Expands while freezing

Vedclass · Answer

(A) According to the Clausius-Clapeyron equation,the change in freezing point with pressure is given by $\frac{dT}{dP} = \frac{T(V_2 - V_1)}{L}$,where $T$ is the temperature,$V_2$ and $V_1$ are the volumes of the solid and liquid phases respectively,and $L$ is the latent heat of fusion.If the freezing point decreases with an increase in pressure,then $\frac{dT}{dP} Since $T$ and $L$ are positive,this implies that $(V_2 - V_1) This indicates that the volume of the solid is less than the volume of the liquid,meaning the substance expands upon melting or contracts upon freezing.However,for substances like water,which expand upon freezing $(V_{solid} > V_{liquid})$,the freezing point decreases with increasing pressure.Therefore,the correct condition is that the liquid expands while freezing.

The freezing point of a liquid decreases when pressure is increased,if the liquid:

Similar Questions

Two different metal wires are joined to form a thermocouple. One junction is kept at a constant temperature,while the other junction is heated to a higher temperature. Which graph represents the variation of thermo $e.m.f.$ $(E)$ with temperature $(t)$?

Find the quantity of heat required to convert $40 \; g$ of ice at $-20^{\circ} C$ into water at $20^{\circ} C$. Given $L_{\text{ice}} = 0.336 \times 10^6 \; J/kg$,specific heat of ice $= 2100 \; J/kg \cdot K$,and specific heat of water $= 4200 \; J/kg \cdot K$. (in $; J$)

The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are $8 \Omega$ and $10 \Omega$ respectively. After inserting in a hot bath of temperature $400^{\circ} C$,the resistance of the platinum wire is: (in $Omega$)

$5 \, g$ of ice at $0^{\circ}C$ is mixed with $5 \, g$ of steam at $100^{\circ}C$. The final temperature of the mixture is ... $^{\circ}C$.

The heat required to convert $1 \, g$ of ice at $0^{\circ}C$ into steam at $100^{\circ}C$ is (given $L_{f} = 80 \, cal/g$,$c_{w} = 1 \, cal/g^{\circ}C$,$L_{v} = 536 \, cal/g$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module