Question

Find the heat produced from an 8.00 L cylinder of propane gas under 5.00 atm at 25.0 oC, if one mole of propane can produce 2220 kJ.

A. 4290 kJ
B. 0.0289 kJ
C. 877 kJ
D. 1.63 kJ
E. 5420 kJ
F. 1750 kJ
G. 8440 kJ
H. 1360 kJ
I. 37.2 kJ
J. 630 kJ
K. 266 kJ
L. 645 kJ
M. 2420 kJ
N. 7.36 x 10-4 kJ

Answer 1

Answer:  3597 kJ of heat

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 5.00 atm

V = Volume of gas = 8.00 L

n = number of moles = ?

R = gas constant =$0.0821Latm/Kmol$

T =temperature =$25.0^0C=(25.0+273)K=298K$

$n=\frac{PV}{RT}$

$n=\frac{5.00atm\times 8.00L}{0.0821 L atm/K mol\times 298K}=1.63moles$

As it is given :

1 mole of propane produces = 2220 kJ of heat

Thus 1.63 moles of propane produces = $\frac{2200}{1}\times 1.63=3597kJ$

Thus 3597 kJ of heat is produced