That would be 0.26 liters
Hope it help!
~Mqddie
To determine the upper bond
Ec(u) = EmVm + EpVp
Em is the elastic modulus of cobalt.
E₁ is the elastic modulus of the particulate
Vm is the volume fraction of cobalt
Vp is the volume fraction of particulate
substitute
Ec(u) = 200 (Vm) + 700 (Vp)
To determine the lower bound
Ec (l) = EmEp/VmEp+ VpEm
Substitute
Ec (l) = 200(700)/Vm(700) + Vp (200)
Ec (l) = 1400/7Vm+2Vp
Answer:
In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
Explanation:
hope this helps
Explanation:
The given data is as follows.
n = 1 mol, 
Q = 1500 J, R = 8.314 J/mol k
(a) 
And, according to the first law of thermodynamics

And, in an isothermal process the change in internal energy of the gas is zero.
Hence, 0 = Q - W
or, W = Q
Expression for work done in an isothermal process is as follows.
W = 
As W = Q, Hence expression for Q will also be given as follows.
Q = 
Now,

[/tex]\Delta S = nR ln \frac{V_{f}}{V_{i}}[/tex]
= 
= nR ln 2
= 
= 5.76 J/K
Therefore, change in entropy is 5.76 J/K.
(b) As, Q = 
= 
= nRT ln 2
T = 
= 
= 260.4 K
Therefore, temperature of the gas is 260.4 K.