Sarah needs to create an architectural drawing for a museum building with an inclined surface. Which presentation view will be t
1 answer:
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Answer:
For SGID you type this
$ find . -perm /4000
For SUID you type this
$ find . -perm /2000
Explanation:
Auxiliary file permissions, that are commonly referred to as “special permissions” in Linux are needed in order to easily find files which have SUID (Setuid) and SGID (Setgid) set.
After typing
$ find directory -perm /permissions
Then type the commands in the attachment below to obtain a list of these files with SGID and SUID.
Answer:
Ts =Ta E)- 300(
569.5 K
5
Q12-W12 = -4014.26
Mol
AU2s = Q23= 5601.55
Mol
AUs¡ = Ws¡ = -5601.55
Explanation:
A clear details for the question is also attached.
(b) The P,V and T for state 1,2 and 3
P =1 bar Ti = 300 K and Vi from ideal gas Vi=
10
24.9x10 m
=
P-5 bar
Due to step 12 is isothermal: T1 = T2= 300 K and
VVi24.9 x 10x-4.9 x 10-3 *
The values at 3 calclated by Uing step 3l Adiabatic process
B-P ()
Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=
14
Ps-1x(4.99 x 103
P-1x(29x 10)
9.49 barr
And Ts =Ta E)- 300(
569.5 K
5
(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and
Work done for Isotermal process define as
8.314 x 300 In =4014.26
Wi2= RTi ln
mol
And fromn first law of thermodynamic
AU12= W12 +Q12
Q12-W12 = -4014.26
Mol
F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and
Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-
AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53
Inol
Now from first law of thermodynamic the Q23
AU2s = Q23= 5601.55
Mol
For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0
and
AH
C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18
mol
AU=C, (T¡-T)= x 8.314 (300
-5601.55
569.5)
mol
Now from first law of thermodynamie the Ws1
J
mol
AUs¡ = Ws¡ = -5601.55
