Thermodynamic Processes
1 answer:
Answer:
A. Part a is the attachment
B. total work = 10.4kj
Explanation:
T1 = constant temperature
nRT1 = PaVa = PbVb
We write equation as
5ma = Pa, 5L = Va, Vb = 10L(temperature is doubled)
W1 = 25 ln2
W1 = 25 x 0.693
= 17.327kj
The isochoric expansion has no change in volume. So,
W2 = 0
Isothermal compression
T3 = constant temperature
nRT3 = PcVc = PdVd
Pc = 1mpa Vc = 10L Vd = 5L
= 10x-0.693
= -6.93kj
Isochoric compression has no change in volume. Workdone w4 = 0
Total workdone = w1 + w2 + w3 + w4
= 17.33 + 0 + (-6.93) + 0
= 10.4kj
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Answer:
miu = 0.31
Explanation:
The friction force is defined as the product of the normal force by the coefficient of friction.
where:
f = friction force = 3 [N]
miu = friction coefficient
N = normal force [N]
The normal force on a horizontal surface can be calculated by means of the product of mass by gravity.
where:
m = mass = 1 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now replacing in the equation above, we can find the friction coefficient.