Answer:
a)
b)Work done per kg of air=196.84 KJ/Kg
Explanation:
Given: for air.
We know that
So
(a)
(b)Work for adiabatic process
W=
We know that PV=mRT for ideal gas.
W=
Now by putting values
work per kg of air=
Work w=-196.84 KJ/Kg (Negative sign indicate work given to input.)
So work done per kg of air=196.84 KJ/Kg
Answer:
probability P = 0.32
Explanation:
this is incomplete question
i found complete A manufactures makes integrated circuits that each have a resistance layer with a target thickness of 200 units. A circuit won't work well if this thickness varies too much from the target value. These thickness measurements are approximately normally distributed with a mean of 200 units and a standard deviation of 12 units. A random sample of 17 measurements is selected for a quality inspection. We can assume that the measurements in the sample are independent. What is the probability that the mean thickness in these 16 measurements x is farther than 3 units away from the target value?
solution
we know that Standard error is expess as
Standard error =
Standard error =
Standard error = 3
so here we get Z value for 3 units away are from mean are
mean = -1 and + 1
so here
probability P will be
probability P = P( z < -1 or z > 1)
probability P = 0.1587 + 0.1587
probability P = 0.3174
probability P = 0.32
Given:
mass of water, m = 2000 kg
temperature, T = = 303 K
extacted mass of water = 100 kg
Atmospheric pressure, P = 101.325 kPa
Solution:
a) Using Ideal gas equation:
PV = mT (1)
where,
V = volume
m = mass of water
P = atmospheric pressure
R= Rydberg's constant = 8.314 KJ/K
M = molar mass of water = 18 g/ mol
Now, using eqn (1):
Therefore, the volume of the tank is
b) After extracting 100 kg of water, amount of water left, m' = m - 100
m' = 2000 - 100 = 1900 kg
The remaining water reaches thermal equilibrium with surrounding temperature at T' = = 303 K
At equilibrium, volume remain same
So,
P'V = m'T'
Therefore, the final pressure is P' = 96.258 kPa
