Question

one mole of oxygen is heated at constant
pressure starting at 0°C. The heat energy that
must be supplied to the gas to double its
volume (R is the molar gas constant)
1) 2.5 x 273 x R 2) 3.5 x 273x R
3) 2.5 x 546 x R 4) 3.5x546x R​

Answer 1

Answer:

option B) 3.5 * 273 * R

Explanation:

therefore Cv = 5/2 R.

So, Cp = Cv + R

= 5/2 R +R

= 7/2 R

At constant pressure, if we double the volume, the temperature will be doubled.

Ti = 0° C

= (0+273) K

= 273 K

So, Tf = 2(273 K)

= 546 K

To obtain heat which added to the gas to double its volume, substitute 1 mol for n, 7/2 R for Cp , 546 K for Tf and 284 K for Ti in the equation Q = nCp(Tf –Ti),

Q = nCp(Tf –Ti)

= (1 mol) (7/2 R) (546 K- 273 K)

= (1 mol) (7/2 ×8.31 J/mol. K) (546 K-273 K) (Since, R = 8.31 J/mol. K)

= 3.5* 273*R