Hence,

Q = H = n(5/2)RT

=>

T = Q/( n(5/2)R )

= 43.08J / (3.00410mol (5/2) 8.314472J/molK )

= 69K = 69 C

(a)

Since no work is done in a constant volume process, the change of internal energy equals the heat transferred to the gas.

U = Q

Assuming ideal gas behavior, the internal energy of the argon sample ideal gas is given by:

U = nCvT

thus,

U = nCvT

The molar heat capacity at constant volume of a monatomic ideal gas like argon is:

Cv = (3/2)R

The number of moles in the sample is

n = m/M = 1.2g / 39.948g/mol = 3.00410mol

Hence,

Q = U = n(3/2)RT

= 3.00410mol (3/2) 8.314472J/molK 115K

= 43.08J

(b)

The heat transferred in a constant pressure process equals the change of enthalpy.

H = Q

Internal energy of an ideal gas is given by:

H = nCpT

<=>

H = nCpT

The molar heat capacity at constant pressure of a monatomic ideal gas is:

Cp = (5/2)R

Hence,

Q = H = n(5/2)RT

=>

T = Q/( n(5/2)R )

= 43.08J / (3.00410mol (5/2) 8.314472J/molK )

= 69K = 69 C

This post is last updated on hrtanswers.com at Date : 1st of September – 2022