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?
Given
Mass of Neon gas mN = 4.0 g
Mass of Oxygen gas mO= 12.0 g
Initial temperature of Neon gas TN = 300 K
Initial Temperature of Oxygen gas TO= 600 K
Molar mass of Neon MN = 20 g/mol
Molar mass of Oxygen MO = 16 g/mol
Universal gas constant R = 8.31J/molK
Solution
Number of moles in 4 grams of Neon gas
nN = mN / MN
nN = 4.0/20
nN = 0.2 mol
Number of moles in 12 grams of Oxygen gas
nO = mN / MN
nO = 12.0/20
nO = 0.6 mol
i)
Initial energy of Neon
UN = 3nNRTN/2
UN = 3 x 0.2 x 8.31 x 300 / 2
UN = 747.9 J
Initial energy of Oxygen
UO = 3nORTO/2
UO = 3 x 0.6 x 8.31 x 600 / 2
UO = 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
UN’+UO’ =UN + UO …….(1)
The energy exchange will stop when both the gases are at the same temperature. Let T’ be the final temperature
UN’ = 3nNRT’/2
UO’ = 3nORT’/2
UN’/ UO’ = nN/nO
UN’ = nN UO’ /nO
Substituting these values in equation (1) we get
(nN UO’ /nO) +UO’ =UN + UO
UO’ {( nN /nO)+1} = 747.9 + 4487.4
UO’ {( 0.2/0.6)+1} = 5235.3
UO’ (0.8/0.6) = 5235.3
UO’ = 5235.3 x 0.6 / 0.8
UO’ = 3926.5 J is the final energy of Oxygen
UN’ = nN UO’ /nO
UN’ = 0.2 x 3926.5 /0.6
UN’ = 1308.8 J is the final energy of Neon
iii)
Change in neon gas energy
ΔUN = UN’ - UN
ΔUN = 1308.8 - 747.9
ΔUN = 560.9 J
Since the value of ΔUN is positive the neon gas gained 560.9 J.
Hence 560.9 J energy is transferred from oxygen to neon
iv)
UN’ = 3nNRT’/2
1308.8 = 3 x 0.2 x 8.31 x T’ /2
T’ = 262.5 K is the final temperature
4.0g of neon gas (Ne) at an initial temperature of 300K interacts thermally with 12.0g of...
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