Question

2.3mol of monatomic gas A initially has 4700J of thermal energy. It interacts with 2.6mol of monatomic gas B, which initially has 8500J of thermal energy.
Which gas has the higher initial temperature?
Gas A or B?
1-What is the final thermal energy of the gas A?
2-What is the final thermal energy of the gas B?

Answer 1

Answer:

Gas B has the higher initial temperature

6,199 J

7,008 J

Explanation:

Mathematically;

The thermal energy of a gas is given by:

E = 3/2 n kT

Where n is the number of moles, K is the molar gas constant and T is the temperature

For Gas A;

4700 = 1.5 * 2.3 * 8.31 * T

T = 4700/28.6695

Thus, T = 163.94 K

For gas B

8500 = 1.5 * 2.6 * 8.31 * T

T = 8500/32.409

T = 262.27 K

This means that gas B has a higher temperature than gas A.

At equilibrium, temperature

T = naTa + nbTb / (na + nb )

T = [2.3(163.94) + 2.6(262.27)]/(2.3 + 2.6)

T = [377.062 + 681.902]/4.9 = 216.12 K

216.12 K is the equilibrium temperature

= 216.12 K is the equilibrium temperature.

Thus, final thermal energy of Gas A and B

Gas A = 1.5 * 2.3 * 8.314 * 216.12= 6,199 J

Gas B = 1.5 * 2.6 * 8.314 * 216.12 = 7,008 J