Question

4.0g of neon gas (Ne) at an initial temperature of 300K interacts thermally with 12.0g of oxygen gas (O2) at an initial temperature of 600K. The molar mass of atomic neon is 20g/mol, the molar mass of atomic oxygen is 16g/mol.
A) What is the initial thermal energy of each gas?
B) What is the final thermal energy of each gas?
C) How much heat energy is transferred and in which direction?
D) What is the final temperature?

Answer 1

Given

Mass of Neon gas                                            m_N = 4.0 g

Mass of Oxygen gas                                        m_O= 12.0 g

Initial temperature of Neon gas                                T_N = 300 K

Initial Temperature of Oxygen gas           T_O= 600 K

Molar mass of Neon                                       M­_N = 20 g/mol

Molar mass of Oxygen                                   M_O = 16 g/mol

Universal gas constant                                   R = 8.31J/molK

Solution

Number of moles in 4 grams of Neon gas

n_N = m_N / M­_N

n_N = 4.0/20

n_N = 0.2 mol

Number of moles in 12 grams of Oxygen gas

n_O = m_N / M­_N

n_O = 12.0/20

n_O = 0.6 mol

i)

Initial energy of Neon

U_N = 3n_NRT_N/2

U_N = 3 x 0.2 x 8.31 x 300 / 2

U_N = 747.9 J

Initial energy of Oxygen

U_O = 3n_ORT_O/2

U_O = 3 x 0.6 x 8.31 x 600 / 2

U_O = 4487.4 J

ii)

Using the conservation of energy principle

Final energy of Neon + Final energy of Oxygen = Initial energy of Neon + Initial energy of Oxygen

U_N’+U_O’ =U_N + U_O                             …….(1)                 

The energy exchange will stop when both the gases are at the same temperature. Let T’ be the final temperature

U_N’ = 3n_NRT’/2

U_O’ = 3n_ORT’/2

U_N’/ U_O’ = n_N/n_O

U_N’ = n_N U_O’ /n_O

Substituting these values in equation (1) we get

(n_N U_O’ /n_O) +U_O’ =U_N + U_O

U_O’ {( n_N /n_O)+1} = 747.9 + 4487.4

U_O’ {( 0.2/0.6)+1} = 5235.3

U_O’ (0.8/0.6) = 5235.3

U_O’ = 5235.3 x 0.6 / 0.8

U_O’ = 3926.5 J is the final energy of Oxygen

U_N’ = n_N U_O’ /n_O

U_N’ = 0.2 x 3926.5 /0.6

U_N’ = 1308.8 J is the final energy of Neon

iii)

Change in neon gas energy

ΔU_N = U_N’ - U_N

ΔU_N = 1308.8 - 747.9

ΔU_N = 560.9 J

Since the value of ΔU_N is positive the neon gas gained 560.9 J.

Hence 560.9 J energy is transferred from oxygen to neon

iv)

U_N’ = 3n_NRT’/2

1308.8 = 3 x 0.2 x 8.31 x T’ /2

T’ = 262.5 K is the final temperature