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}
Assume that the universal gas constant is 8314 J \;kmol^{-1}K^{-1}
2 | |
4 | |
6 | |
8 |
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 |
Which one of the following statements is FALSE?
For an ideal gas, the enthalpy is independent of pressure. | |
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. | |
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. | |
Any real gas behaves as an ideal gas at sufficiently low pressure or sufficiently high
temperature. |
Question 2 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.
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 3 |
At steady state, 500 kg/s of steam enters a
turbine with specific enthalpy equal to 3500 kJ/
kg and specific entropy equal to 6.5 kJ \dot kg^{-1}\dot K^{-1}. It
expands reversibly in the turbine to the condenser
pressure. Heat loss occurs reversibly in the turbine
at a temperature of 500 K. If the exit specific
enthalpy and specific entropy are 2500 kJ/kg and
6.3 kJ\dot kg^{-1} \dot K^{-1}, respectively, the work output from
the turbine is ________ MW (in integer).
320 | |
650 | |
275 | |
450 |
Question 3 Explanation:
Mass flow rate of stem \dot{m}=500kg/s
h_1=300 kJ/kg
S_1=6.5 kJ/kgL
h_2=2500kJ/kg
S_2=6.3 kJ/kgK
Surrounding temperature T_o=500 K
Work output from turbine
\begin{aligned} W_T&=(h_1-h_2)-T_o(S_1-S_2)\\ &=(3500-2500)-500(6.5-6.3)\\ &=1000-100\\ &=900 kJ/kg \end{aligned}
Power output of turbine
\begin{aligned} P&=\frac{W_T}{kg} \times \dot{m}\\ &=900 \times 500\\ &=450000 kW\\ &=450 MW \end{aligned}
h_1=300 kJ/kg
S_1=6.5 kJ/kgL
h_2=2500kJ/kg
S_2=6.3 kJ/kgK
Surrounding temperature T_o=500 K
Work output from turbine
\begin{aligned} W_T&=(h_1-h_2)-T_o(S_1-S_2)\\ &=(3500-2500)-500(6.5-6.3)\\ &=1000-100\\ &=900 kJ/kg \end{aligned}
Power output of turbine
\begin{aligned} P&=\frac{W_T}{kg} \times \dot{m}\\ &=900 \times 500\\ &=450000 kW\\ &=450 MW \end{aligned}
Question 4 |
A rigid tank of volume of 8 m^3 is being filled up
with air from a pipeline connected through a valve.
Initially the valve is closed and the tank is assumed
to be completely evacuated. The air pressure and
temperature inside the pipeline are maintained at
600 kPa and 306 K, respectively. The filling of the
tank begins by opening the valve and the process
ends when the tank pressure is equal to the pipeline
pressure. During the filling process, heat loss to
the surrounding is 1000 kJ. The specific heats of
air at constant pressure and at constant volume are
1.005 kJ/kg.K and 0.718 kJ/kg.K, respectively.
Neglect changes in kinetic energy and potential
energy.
The final temperature of the tank after the completion of the filling process is _________ K (round off to the nearest integer).
The final temperature of the tank after the completion of the filling process is _________ K (round off to the nearest integer).
395 | |
254 | |
355 | |
125 |
Question 4 Explanation:
Initially the tank is completely evacuated (m_1 = 0)
After the gas is filled with tank, the gas pressure in the tank is 600 kPa.
Heat lost to surroundings (Q) = 1000 kJ
c_p=1.005 KJ/kgK
c_v=0.718KJ/kgK
R=c_p-c_v=0.287 kJ/kgK
(Changes in KE and PE are negligible)
Final temperature of gas T_f=?
m = mass of gas entering to tank
The energy balance equation is energy carried by gas in the pipe = energy of gas in the rigid tank + Heat lost to surrounding
\begin{aligned} mh_i&=mu_f+Q\\ h_i&=u_f+\frac{Q}{m}\\ C+pT_i&=C_vT_f+\frac{1000}{\frac{16724.73868}{T_f}} \end{aligned}
Mass of gas in the tank
\begin{aligned} m=\frac{P_fV_f}{RT_f}\\ &=\frac{600(8)}{0.287(T_f)}\\ &=\frac{16724.73868}{T_f}\\ 1.005(306)&=0.718T_f+\frac{1000T_f}{16724.73868}\\ &=0.718T_f+0.05979T_f\\ &=0.77779T_f\\ \therefore \; T_f&=\frac{1.005 \times 306}{0.77779}\\ &=395.389K\approx 395K \end{aligned}
Question 5 |
Which one of the following is an intensive property
of a thermodynamic system?
Mass | |
Density | |
Energy | |
Volume |
Question 5 Explanation:
Intensive property -> Density
Mass, energy and volume are extensive properties.
Mass, energy and volume are extensive properties.
There are 5 questions to complete.
great work,user friendly site, but one suggestion for some questions (like q1) ,please add some explanation like formula name or concept usedlike that 👍👏
Awesome👏
Website
In qu-8:
Given, Patm = 1bar =100kpa and Pgauge = 0.5bar = 50kpa,V = 2.87m3,T = 300K
so, Pabs = Patm + Pgauge = 100+50 = 150kpa
Therefore, Mair = (150*1000*2.87)/(287*300) = 5kg
Very helpful