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
Answer:
plasmodium belongs to kingdom protista
In order to calculate the amount of energy required, we must first check the latent heat of vaporization of water from literature. The latent heat of vaporization of any substance is the amount of energy required per unit mass to convert that substance from a solid to a liquid. For water this is 2,260 J/g. We now use the formula:
Energy = mass * latent heat
Q = 50 * 2,260
Q = 113,000 J
113,000 Joules of heat energy are required.