FUND OF ENG THERMODYN(LLF)+WILEYPLUS
9th Edition
ISBN: 9781119391777
Author: MORAN
Publisher: WILEY
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1. Complete the following table. Show the states on the T-v diagram
Substance
T°C
P, KPa
Internal Energy.
Enthalpy,
Specific volume,
Quality, X
Phase description
u, kj/kg
h, kj/kg
v, m'/kg
Pg 98 pa
H,0
200
10000
H,O
600
2000
R-134a
900
180
Q1: a) If for a given gas (Ou/ov)0, find u, the specific internal
energy if c, is constant.
b) If the equation of state for the gas in part (a) is given by
v-(RT/p) = a, find c,- c, for that gas.
c) Show that (@h/op)r=-C, (@T/ôp)n.
One mole of gas undergoes the cycle shown in the p-V diagram below.
A
Po
B
1.
D
Po/32
Vo 2Vo
8V
Volume
16V.
(a) Classify each process in this cycle: A-B, B-C, C-D, and D-A.
(b) Find y for this gas.
(c) What type of gas is this, monatomic, diatomic, or polyatomic?
Pressure
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- (5) Consider a gas that follows the equation of state NRT aN? p = V – bN V2 | 8a This is called van der Waals gas (a,b > 0). For temperatures T > Tc = p(V) is a 27RB monotonic decreasing function. Assume T > Tc and Cy = 3NR/2. Consider an insulated box with volume V, . The box was divided by a wall. One of the compartments had volume Vo at temperature To, filled with van der Waals gas of amount N, and the other was vacuum. The system was in equilibrium. (a) The wall was removed abruptly. The gas expanded and occupied the entire box. This process is called adiabatic free expansion. The temperature in the box is now T, in equilibrium. What is T, – T,? (b) The internal energy of the gas does not change by adiabatic free expansion. Why? (c) Instead of the abrupt removal, the wall was moved through adiabatic and quasi-static process, and the gas expanded to the entire volume of the box. The temperature in the box is now T, in equilibrium. What is T2? (d) Show T, > T2, using the arguments of…arrow_forward2. Given a closed container of volume 1 L containing 50 g of helium gas, calculate the following: a. The pressure at T = 6 K, assuming that the gas is an ideal gas. b. The pressure at T of state. = 6 K, assuming that the gas obeys van der Waal's equation c. The pressure at T 6 K, assuming that the gas obeys the virial equation of state truncated after the second term. = d. The state of the fluid at T = 4.2 K. (Hint: 4.2 K is below T. and 50 kg/m³ is greater than p, @ 4.2 K.)arrow_forward2) Using the tables for water, determine the specified property data at the indicated states. In each case, locate the state on a sketch of the p–v diagram. a)At p = 3 bar, v= 0.5 m3/kg, find T in °C and u in kJ/kg. b)At T = 320°C, v= 0.03 m3/kg, find p in MPa and u in kJ/kg. c)At p = 28 MPa, T = 520°C, find in m3/kg and h in kJ/kg.arrow_forward
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