Thermodynamics


Question 1
Consider a mixture of two ideal gases, X and Y, with molar masses \bar{M}_X=10kg/kmol and \bar{M}_Y=20kg/kmol , respectively, in a container. The total pressure in the container is 100 kPa, the total volume of the container is 10m^3 and the temperature of the contents of the container is 300 K. If the mass of gas-X in the container is 2 kg, then the mass of gas-Y in the container is ____ kg. (Rounded off to one decimal place)
Assume that the universal gas constant is 8314 J \;kmol^{-1}K^{-1}
A
2
B
4
C
6
D
8
GATE ME 2023      Thermodynamic System and Processes
Question 1 Explanation: 


By gas law
\mathrm{Pv}=\mathrm{nRT}
where \mathrm{n}= total value of gas in container
\begin{aligned} 100 \times 10 & =n \times 8.314 \times 300 \\ n & =0.4 \end{aligned}
and \mathrm{n}=\mathrm{n}_{\mathrm{x}}+\mathrm{n}_{\mathrm{y}}
Or, 0.4=\frac{m_{x}}{M_{x}}+\frac{m_{y}}{M_{y}}=\frac{2}{10}+\frac{m_{y}}{20} or, m_{y}=4 \mathrm{~kg}
Question 2
Consider a fully adiabatic piston-cylinder arrangement as shown in the figure. The piston is massless and cross-sectional area of the cylinder is A. The fluid inside the cylinder is air (considered as a perfect gas), with \gamma being the ratio of the specific heat at constant pressure to the specific heat at constant volume for air. The piston is initially located at a position L_1 . The initial pressure of the air inside the cylinder is P_1 \gt \gt P_0 , where P_0 is the atmospheric pressure. The stop S_1 is instantaneously removed and the piston moves to the position L_2 , where the equilibrium pressure of air inside the cylinder is P_2 \gt \gt P_0 .
What is the work done by the piston on the atmosphere during this process?

A
0
B
P_0A(L_2-L_1)
C
P_1AL_1 \ln \frac{L_1}{L_2}
D
\frac{(P_2L_2-P_1L_1)A}{(1-\gamma )}
GATE ME 2023      First Law, Heat, Work and Energy
Question 2 Explanation: 


Initial volume \mathrm{V}_{1}=\mathrm{L}_{1} \times \mathrm{A}
Final volume V_{2}=L_{2} \times A
Work done by atmospheric air =\int_{V_{1}}^{V_{2}} P d v
\begin{aligned} & =P_{0} \int_{V_{1}}^{V_{2}} d V \\ & =P_{0}\left(L_{2} A-L_{1} A\right) \\ & =P_{0} A\left(L_{2}-L_{1}\right) \end{aligned}


Question 3
Which one of the following statements is FALSE?
A
For an ideal gas, the enthalpy is independent of pressure.
B
For a real gas going through an adiabatic reversible process, the process equation is given by PV^\gamma = constant, where P is the pressure, V is the volume and \gamma is the ratio of the specific heats of the gas at constant pressure and constant volume.
C
For an ideal gas undergoing a reversible polytropic process PV^{1.5} = constant, the equation connecting the pressure, volume and temperature of the gas at any point along the process is \frac{P}{R}=\frac{mT}{V} where R is the gas constant and m is the mass of the gas.
D
Any real gas behaves as an ideal gas at sufficiently low pressure or sufficiently high temperature.
GATE ME 2023      Thermodynamic System and Processes
Question 3 Explanation: 
Ideal gas follow the adiabatic equation \mathrm{PV}^{\gamma}=\mathrm{C} and also follow the gas law P V=m R T
Any real gas at low pressure and high temperature follow gas law, e.g. Air conditioning system.
Real gas follow Van der waal's gas equation for an adiabatic process as
\left(P+\frac{n^{2} a}{V^{a}}\right)(V-n b)^{\top}=k
where
\begin{aligned} & \mathrm{T}=\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}+1 \\ & \mathrm{k}=\mathrm{nRZ} \end{aligned}
Enthalpy of an ideal gas is independent of pressure at constant temperature and the internal energy of an ideal gas is independent of volume at constant temperature.
Question 4
A heat engine extracts heat (Q_H) from a thermal reservoir at a temperature of 1000 K and rejects heat (Q_L) to a thermal reservoir at a temperature of 100 K, while producing work (W). Which one of the combinations of (Q_H,Q_L,W) given is allowed?
A
Q_H=2000J, Q_L=500J, W=1000J
B
Q_H=2000J, Q_L=750J, W=1250J
C
Q_H=6000J, Q_L=500J, W=5500J
D
Q_H=6000J, Q_L=600J, W=5500J
GATE ME 2023      First Law, Heat, Work and Energy
Question 4 Explanation: 
For a reversible engine, the rate of heat rejection is minimum.

For process to be feasible
\oint \frac{dQ}{T}\leq 0
for option (b) \oint \frac{dQ}{T}=\frac{Q_H}{T_1}-\frac{Q_2}{T_2}=\frac{2000}{1000}-\frac{750}{100} \lt 0
So cyclic process is possible.
Question 5
Consider an isentropic flow of air (ratio of specific heats = 1.4) through a duct as shown in the figure.
The variations in the flow across the cross-section are negligible. The flow conditions at Location 1 are given as follows:

?? P_1=100kPa, \rho _1=1.2kg/m^3,u_1=400m/s

The duct cross-sectional area at Location 2 is given by A_2=2A_1, where A_1 denotes the duct cross-sectional area at Location 1. Which one of the given statements about the velocity u_2 and pressure P_2 at Location 2 is TRUE?

A
u_2 \lt u_1, P_2 \lt P_1
B
u_2 \lt u_1, P_2 \gt P_1
C
u_2 \gt u_1, P_2 \lt P_1
D
u_2 \gt u_1, P_2 \gt P_1
GATE ME 2023      Power System
Question 5 Explanation: 
Step -1: First identify type of flow - Subsonic or Supersonic by finding out Mach number

Mach no at start of flow
M_a=\frac{u_1}{C_1},\; where, \; C_1=\sqrt{\gamma RT_1}
u_1 is velocity of gas
C_1 is velocity of sound

\begin{aligned} P_1 &=\rho _1RT_1 \\ T_1&= \frac{P_1}{\rho _1R}=\frac{100}{1.2 \times 0.287}=290.36K\\ C_1 &=\sqrt{1.4 \times 287 \times 290.36} \\ &= 341.56 m/sec\\ Ma_1 &=\frac{400}{341.56}=1.017 \end{aligned}
Flow is supersonic flow. Hence, diverging duct is nozzle so u_2 \gt u_1 and P_2 \lt P_1 .




There are 5 questions to complete.

2 thoughts on “Thermodynamics”

Leave a Comment