Thermodynamics Test

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Thermodynamics Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) to (4.0 atm, 5.0 L, 245 K) with a change in internal energy, ∆U = 30 L atm. The change in enthalpy (∆H) of the process is

A
30.0 L atm

B
40.0 L atm

C
44.0 L atm

D
56.0 L atm

Solution

∆H = ∆U + ∆(PV) = ∆U + (P₂V₂ - P₁V₁)
∆H = 30.0 L atm + (4.0 atm × 5.0 L) - (2.0 atm × 3.0 L) = 44.0 L atm
Question 2
1 / -0

For which of the following equations is ∆_rH equal to ∆_fH?

A
CH₄(g) + Cl₂(g) → CH₂Cl₂(l) + 2HCl (g)

B
Xe (g) + 2F₂(g) → XeF₄(g)

C
2CO (g) + O₂(g) → 2CO₂(g)

D
N₂ (g) + O₃(g) → N₂O₃(g)

Solution

∆_fH is the amount of heat evolved when one mole of a compound is synthesised from its constituent elements in their standard states.
The equation given under option 2 represents the formation of one mole of XeF₄ from its elements in their standard states; and hence, the heat of reaction, ∆_rH, equals the heat of formation, ∆_fH of XeF₄.
NOTE: Option 4 also represents the formation of 1 mole of N₂O₃ from its constituent elements, but the standard state of N₂O₃ is not gaseous.
Question 3
1 / -0

A balloon filled with a certain quantity of a gas expands its volume by 2.0 L. If the pressure outside the balloon is 0.93 bar and the energy change of the gas is 450 J, how much heat did the surroundings give to the balloon?

A
636 J

B
-636 J

C
400 J

D
-400 J

Solution

According to the first law of thermodynamics,
U = q + w
U = 450 J
Expansion work done = -PV = -0.93 x 2.0 = -1.86 bar-L = -1.86 x 100 = -186 J
Therefore, q = U - w
q = 450 - (-186) = 636 J
Question 4
1 / -0

If a monatomic ideal gas is undergoing a process in which the ratio of P to V at any instant is constant and is equal to 1, then the molar heat capacity of the gas at constant volume is

A
3R/2

B
2R

C
5R/2

D
3R

Solution

From first law of thermodynamics,
DE = q + w
nC_vdT = nC_pdT – PdV ... (1)
Now according to process, P = V
And according to ideal gas equation, PV = nRT
We have, V² = nRT
On differentiating, we get
2VdV = nRdT and PdV = VdV = nRdT/2
So, from equation (1), we have
nC_vdT = nC_pdT – nRdT/ 2
So, C_v = C_p – R/2
For a monatomic gas, C_p = 5/2 R

Therefore, C_v = 5/2 R - 1/2 R = 4/2 R or 2R