A sample of argon of 1 atm pressure and 300K expands reversibly and adiabatically from `1.25 dm^(3) " to " 2.5 dm^(3)`. Calculate the approximate enthalpy (in J) change
(i) `C_(V)` for argon is `12.48 JK^(-1)`
(ii) Assume argon to be an ideal gas
(iii) `Delta T= 111.5K`

Options

A. 20.9
B. 117
C. 234
D. 58.5

Answer

B

Explanation

We know that, `Delta H = nC_(p) Delta T`
Number of moles of argon `(n) = (pV)/(RT)`
Given, p = 1 atm, `V = 1.25 dm^(3)`, T = 300 K
`:. n = (1 xx 1.25)/(0.082 xx 300) = 0.05` mol
Also, `C_(p) C_(V) = 12.48JK^(-1)`
`C_(p) = C_(V) + R = 1248 + 8.314 = 20.794`
On substituting the above values in Eq (i) , we get
`Delta H = 0.05 xx 20.794 xx 111.5 = 115.92 ~~ 117J`