Consider the phase diagram of a one-component system given below. \( V_{\alpha} \), \( V_{\beta} \), and \( V_{{Liquid}} \) are the molar volumes of \( \alpha \), \( \beta \), and liquid phases, respectively. Which one of the following statements is TRUE? Given: The change in molar enthalpies, \( \Delta H_{\alpha \to \beta} \) and \( \Delta H_{\beta \to {Liquid}} \), are positive.
In the phase diagram, the key information provided is that the enthalpy changes for the transitions \( \alpha \to \beta \) and \( \beta \to {Liquid} \) are positive. This suggests that both transitions require heat absorption, which typically corresponds to a change from a denser phase to a less dense phase. Let's analyze the phases and their associated molar volumes:
- The transition from \( \alpha \) to \( \beta \) involves an increase in volume, so \( V_{\beta}>V_{\alpha} \).
- The transition from \( \beta \) to liquid also involves an increase in volume, so \( V_{{Liquid}}>V_{\beta} \).
Thus, the correct relationship between the molar volumes is: \[ V_{\alpha}<V_{\beta}<V_{{Liquid}} \] This matches Option B.
Step 1: Analysis of each option - Option (A): \( V_{\alpha}<V_{\beta} \) and \( V_{\beta}<V_{{Liquid}} \) - Incorrect: While \( V_{\alpha}<V_{\beta} \) is correct, the condition \( V_{\beta}<V_{{Liquid}} \) does not align with the enthalpy changes. The liquid phase has a larger volume than the \( \beta \)-phase, so this option is incorrect.
- Option (B): \( V_{\alpha}>V_{\beta} \) and \( V_{\beta}<V_{{Liquid}} \) - Correct: This matches the expected sequence of molar volumes based on the phase transitions and the positive enthalpy changes. The volume of the liquid phase is greater than that of the \( \beta \)-phase, which is larger than that of the \( \alpha \)-phase.
- Option (C): \( V_{\alpha}<V_{\beta} \) and \( V_{\beta}>V_{{Liquid}} \) - Incorrect: This contradicts the typical behavior of phase transitions. The liquid phase should have a larger volume than the \( \beta \)-phase, making this option incorrect.
- Option (D): \( V_{\alpha}>V_{\beta} \) and \( V_{\beta}>V_{{Liquid}} \) - Incorrect: This option suggests that the \( \beta \)-phase has a larger volume than the liquid phase, which is not true. Hence, this option is also incorrect.
Step 2: Conclusion The correct option is Option B, where \( V_{\alpha}>V_{\beta} \) and \( V_{\beta}<V_{{Liquid}} \). This corresponds to the expected volume changes based on the enthalpy and phase transitions.
Radiative heat flux \( \dot{q} \) at a hot surface at a temperature \( T_s \) can be expressed as \[ \dot{q} = A f(T_s, T_\infty) (T_s - T_\infty) \] where \( A \) is a constant and \( T_\infty \) is the temperature of the surroundings (temperatures are expressed in K). The function \( f(T_s, T_\infty) \) is given by ______.
Match the steel plant related processes in Column I with the associated information in Column II.
For two continuous functions \( M(x, y) \) and \( N(x, y) \), the relation \( M dx + N dy = 0 \) describes an exact differential equation if
A linear regression model was fitted to a set of \( (x, y) \) data. The total sum of squares and sum of squares of error are 1200 and 120, respectively. The coefficient of determination \( R^2 \) of the fit is ......... (rounded off to one decimal place).
The hydrostatic stress for the stress tensor provided below is ......... MPa (in integer). \[ \begin{bmatrix} 150 & 0 & 0 \\ 0 & -100 & 100 \\ 0 & 100 & 250 \end{bmatrix} \, \text{MPa} \]
Match the phenomena in Column I with the typical observations in Column II.
Consider the gas phase reaction: \[ CO + \frac{1}{2} O_2 \rightleftharpoons CO_2 \] At equilibrium for a particular temperature, the partial pressures of \( CO \), \( O_2 \), and \( CO_2 \) are found to be \( 10^{-6} \, {atm} \), \( 10^{-6} \, {atm} \), and \( 16 \, {atm} \), respectively. The equilibrium constant for the reaction is ......... \( \times 10^{10} \) (rounded off to one decimal place).
For an application where the Reynolds number is to be kept constant, a liquid with a density of 1 g cm\(^-3\) and viscosity of 0.01 Poise results in a characteristic speed of 1 cm s\(^-1\). If this liquid is replaced by another with a density of 1.25 g cm\(^-3\) and viscosity of 0.015 Poise, the characteristic velocity will be ......... cm s\(^-1\) (rounded off to one decimal place).
Consider a fully developed, steady, one-dimensional, laminar flow of a Newtonian liquid through a pipe. The maximum velocity in the pipe is proportional to which of the following quantities?
Given: \( \Delta P \) is the difference between the outlet and inlet pressure, \( \mu \) is the dynamic viscosity of the liquid, and \( R \) and \( L \) are the radius and length of the pipe, respectively.