One kmol of an ideal gas undergoes the following reversible cycle in a closed system: - Step 1ĺ2: Isothermal expansion at T = 600 K from 5 to 4 bar - Step 2ĺ3: Adiabatic expansion from 4 to 3 bar - Step 3ĺ4: Isobaric cooling to T = 360 K - Step 4ĺ1: Isochoric compression to return to the initial state Given: Cp = 30.0 J/(mol. K) and R = 8.314 J/(mol. K). a) Determine the pressure (in bar), the temperature (in K) and the volume (in m3 ) of the ideal gas at every stage in the cycle b) Calculate Q, W, ǻU and ǻH for each step of the cycle c) Calculate Q, W, ǻU and ǻH for each step of the cycle in case the first three steps are irreversible with an efficiency of 75% compared to the reversible process. Assume that all changes of state are identical to a) and note that the second step can no longer be adiabatic.

i believe it is-

d.) gravity pulled particles of dust and gas together to form planets.

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

i would say b

explanation:

but i think i would need more information than that

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally