2.50 mole sample of an ideal gas, for which , is subjected to two successive changes in state: (1) from 25.0°c and , the gas is expanded isothermally against a constant pressure of to twice the initial volume. (2) at the end of the previous process, the gas is cooled at constant volume from 25.0°c to –29.0°c. calculate q, w, , and for each of the stages. also calculate q, w, , and for the complete process

(1) W = - 4293.3 J; Q = 4293.3 J

(2) W = 0 J; Q = - 1683.585 J

(3) Wt = -4293.3 J; Qt = 2609.72 J

Explanation:

(1) ΔU = 0

⇒ Q = - W

⇒ W = - ∫PdV

∴P = nRT/V

⇒ W = - nRT ∫dV/V

⇒ W = - nRT Ln (Vf/Vi) = - nRT Ln(2Vi/Vi) = - nRT Ln(2)

⇒ W = - (2.5)(8.314)(298)(0.693) = - 4293.3 J

⇒ Q = 4293.3 J

(2) W = 0

⇒ ΔU = Q = Cv*ΔT

∴ ΔT = 244 - 298 = - 54K

∴ Cv = 1.5*R = 12.471 J/mol.K

⇒ Q = (12.471)*(- 54)*(2.5) = - 1683.585 J

(3) ⇒ Wt = W(1) + W(2) = -4293.3 + 0 = - 4293.3 J

⇒ Qt = Q(1) + Q(2) = 4293.3 - 1683.585 = 2609.72 J

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

the formula for the conversion of temperature from degree celsius to kelvin is:

the sample of water at is converted to kelvin temperature by using the above formula as:

on comparing the two samples of water at or we can say that at and at respectively. the difference between the temperatures of two samples of water is:

hence, one can say that the sample of water at is greater than the sample of water at by .