# Pure Substances

 Question 1
In the vicinity of the triple point, the equation of liquid-vapour boundary in the $P-T$ phase diagram for ammonia is $\ln P=24.38-3063/T$, where $P$ is pressure (in Pa) and $T$ is temperature (in K). Similarly, the solid-vapour boundary is given by $\ln P=27.92-3754/T$. The temperature at the triple point is ________K (round off to one decimal place).
 A 195.2 B 25.6 C 254.6 D 125.5
GATE ME 2021 SET-1   Thermodynamics
Question 1 Explanation: Liquid vapour, $\ln P=24.38-\frac{3063}{T}$
Solid vapour,$\ln P=27.92-\frac{3754}{T}$
At triple point temperature of solid, liquid and vapour is same.
$\therefore$equating
$24.38-\frac{3063}{T}=27.92-\frac{3754}{T}$
Multiplying by T on both sides
\begin{aligned} 24.38 T-3063 &=27.92 T-3754 \\ 3.54 T &=691 \\ T &=195.197 \mathrm{~K} \end{aligned}
 Question 2
A closed vessel contains pure water, in thermal equilibrium with its vapour at 25$^{\circ}C$ (Stage #1), as shown. The vessel in this stage is then kept inside an isothermal oven which is having an atmosphere of hot air maintained at 80$^{\circ}C$. The vessel exchanges heat with the oven atmosphere and attains a new thermal equilibrium (Stage #2). If the Valve A is now opened inside the oven, what will happen immediately after opening the valve?
 A Water vapor inside the vessel will come out of the Valve A B Hot air will go inside the vessel through Valve A C Nothing will happen - the vessel will continue to remain in equilibrium D All the vapor inside the vessel will immediately condense
GATE ME 2020 SET-2   Thermodynamics
Question 2 Explanation:
As inside pressure is less than outside pressure, hot air will ?ow into the vessel.
 Question 3
For an ideal Rankine cycle operating between pressures of 30 bar and 0.04 bar, the work output from the turbine is 903 kJ/kg and the work input to the feed pump is 3 kJ/kg. The specific steam consumption is _________________ kg/kW.h (round off to 2 decimal places).
 A 2 B 3 C 4 D 5
GATE ME 2020 SET-1   Thermodynamics
Question 3 Explanation:
\begin{aligned} (WD)_{\text {turbine }} &=903 \mathrm{kJ} / \mathrm{kg} \\ (WD)_{\text {pump}} &=3 \mathrm{kJ} / \mathrm{kg} \\ \text { Specific steam consumption }=& ?(\mathrm{kg} / \mathrm{kW}-\mathrm{hr}) \\ \mathrm{SSC} &=\frac{3600}{W_{T}-W_{C}} \\ &=\frac{3600}{903-3}=4 \mathrm{kg} / \mathrm{kW}-\mathrm{hr} \end{aligned}
 Question 4
Air (ideal gas) enters a perfectly insulated compressor at a temperature of 310 K. The pressure ratio of the compressor is 6. Specific heat at constant pressure for air is 1005 J/kg.K and ratio of specific heats at constant pressure and constant volume is 1.4. Assume that specific heats of air are constant. If the isentropic efficiency of the compressor is 85 percent, the difference in enthalpies of air between the exit and the inlet of the compressor is ________ kJ/kg (round off to nearest integer).
 A 245 B 264 C 312 D 224
GATE ME 2020 SET-1   Thermodynamics
Question 4 Explanation: \begin{aligned} T_{1} &=310 \mathrm{K} \\ r_{p} &=\frac{P_{2}}{P_{1}}=6 \quad \eta_{s}=0.85 \\ C_{p} &=1.005 \mathrm{kJ} / \mathrm{kgK} \\ \gamma &=1.4 \\ \therefore \quad \frac{T_{2}}{T_{1}} &=\left(\frac{P_{2}}{P_{1}}\right)^{\frac{\gamma-1}{\gamma}} \\ \Rightarrow \quad \frac{T_{2}}{310} &=(6)^{1.4-1 / 1.4} \\ \Rightarrow \quad T_{2} &=517.225 \mathrm{K} \\ \text{Now,}\quad \eta_{\text {isen }} &=\frac{W_{\text {isen }}}{W_{\text {actual }}}=\frac{h_{2}-h_{1}}{h_{2}-h_{1}} \\ 0.85 &=\frac{C_{p}\left(T_{2}-T_{1}\right)}{h_{2}-h_{1}} \\ \Rightarrow \quad h_{2}-h_{1} &=\frac{1.005(517.22-310)}{0.85}=245 \mathrm{kJ} / \mathrm{kg} \end{aligned}
 Question 5
The compressor of a gas turbine plant, operating on an ideal intercooled Brayton cycle, accomplishes an overall compression ratio of 6 in a two-stage compression process, Intercooling is used to cool the air coming out from the first stage to the inlet temperature of the first stage, before its entry to the second stage. Air enters the compressor at 300 K and 100 kPa. If the properties of gas are constant, the intercooling pressure for minimum compressor work is________ kPa (round off to 2 decimal places).
 A 600 B 124.63 C 312.26 D 244.94
GATE ME 2020 SET-1   Thermodynamics
Question 5 Explanation:
$\frac{P_{3}}{P_{1}}=6\, \, \Rightarrow P_{3}=600\, kPa$
intermediate Pressure $=\sqrt{P_{3}P_{1}}=\sqrt{600\times 100}=244.94kPa$
 Question 6
For an ideal gas, a constant pressure line and a constant volume line intersect at a point, in the Temperature (T) versus specific entropy (s) diagram. $C_P$ is the specific heat at constant pressure and $C_V$ is the specific heat at constant volume. The ratio of the slopes of the constant pressure and constant volume lines at the point of intersection is
 A $\frac{C_P-C_V}{C_P}$ B $\frac{C_P}{C_V}$ C $\frac{C_P-C_V}{C_V}$ D $\frac{C_V}{C_P}$
GATE ME 2020 SET-1   Thermodynamics
Question 6 Explanation: \begin{aligned} \text { If } \qquad v & =C \\ T d s & =d u+p d v \\ T d s & =d u\\ d u &=C_{V} d T &[\therefore V=C ]\\ T d s &=C_{V} d T \\ \left(\frac{d T}{d s}\right)_{V} &=\frac{T}{C_{V}} \\ \text{If}\qquad P &=C\\ T d s&=d h-v d P \\ d h &=C_{P} d T \qquad [\therefore P=C]\\ T d s &=C_{P} d T \\ \left(\frac{d T}{d s}\right)_{P} &=\frac{T}{C_{P}} \\ \text{Ratio, }\quad\frac{(d T / d s)_{P}}{(d T / d s)_{V}} &=\frac{\left(T / C_{P}\right)_{P}}{\left(T / C_{V}\right)_{V}}=\frac{C_{V}}{C_{P}} \end{aligned}
 Question 7
For an ideal gas, the value of the Joule-Thomson coefficient is
 A Positive B Negative C Zero D Indeterminate
GATE ME 2020 SET-1   Thermodynamics
Question 7 Explanation:
Value of Joule Thomson coefficient for ideal gas $\mu=\left(\frac{\partial T}{\partial P}\right)_{h}=0$
 Question 8
The INCORRECT statement about the characteristics of critical point of a pure substance is that
 A there is no constant temperature vaporization process B it has point of inflection with zero slope C the ice directly converts from solid phase to vapor phase D saturated liquid and saturated vapor states are identical
GATE ME 2016 SET-3   Thermodynamics
Question 8 Explanation:
At critical point liquid directly converts vapour phase.
 Question 9
For water at $25^{\circ}C$ , $dp_{s}/dT_{s}=0.189 KPa/K$ ($p_{s}$ is the saturation pressure in kPa and $T_{s}$ is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 $m^{3}$ /kg. Assume that the specific volume of liquid is negligible in comparison with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at $25^{\circ}C$ (in kJ/kg) is __________
 A 2444 B 1000 C 1050 D 2000
GATE ME 2016 SET-1   Thermodynamics
Question 9 Explanation:
At $25^{\circ}C, \frac{dp_s}{dT_s}=0.189kPa/K$
Specific volume of saturated vapour,
$v_s=43.38 m^3/kg$
From clausis-clapeyron equation neglecting specific volume of liquid
\begin{aligned} \frac{dp_s}{dT_s}&=\frac{h_{fg}}{T_s \times v_s}\\ h_{fg}&=\text{Enthalpy of vapourisation}\\ 0.189&=\frac{h_{fg}}{(25+273) \times 43.38}\\ \therefore \;\; h_{fg}&=2443.248\; kJ/kg \end{aligned}
 Question 10
A rigid container of volume 0.5 $m^{3}$ contains 1.0 kg of water at $120^{\circ}C$ ($v_{f}$ = 0.00106 $m^{3}$/kg, $v_{g}$= 0.8908 $m^{3}$/kg). The state of water is
 A compressed liquid B saturated liquid C a mixture of saturated liquid and saturated vapor D superheated vapor
GATE ME 2015 SET-3   Thermodynamics
Question 10 Explanation:
$v=\frac{V}{m}=\frac{0.5}{1}=0.5 \mathrm{m}^{3} / \mathrm{kg}$
$\therefore$ It is lying between $v_{f}$ and $v_{g}$. The state of water is a mixture of saturated liquid and saturated vapour.
There are 10 questions to complete.