Thermodynamics Test

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Thermodynamics Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

Statement-1: A process is called adiabatic if the system does not exchange heat with the surroundings.

Statement-2: It does not involve increase or decrease in temperature of the system.

A
Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1.

B
Both Statement-1 and Statement-2 are true and Statement-2 is not the correct explanation of Statement-1.

C
Statement-1 is true but Statement-2 is false.

D
Statement-1 is false but Statement-2 is true.

Solution

It may involve increase or decrease in temperature of the system. Systems in which such processes occur, are thermally insulated from the surroundings.
Question 2
1 / -0

Temperature of 5 moles of a gas is decreased by 2 K at constant pressure. Which of the following statement is correct?

A
work done by the gas = 5 R

B
work done = 0

C
work done by the gas = 10 R

D
work done over the gas = 10 R

Solution

For 5 moles of gas at T, PV₁ = 5RT

For 5 moles of gas at T-2, PV₂ = 5R ( T - 2 )

Hence, PV₂ - PV₁ =P(V₂ - V₁) = PΔV

= 5R [ T - 2 - T ] = - 10R

or, - PΔV = 10R ( ΔV is negative, W is +ve )
Question 3
1 / -0

The enthalpy change for the hydration,

CuSO_4(s) + 5H₂O_(g)→ CuSO₄.5H₂O_(s)

is - 71.50 kcal mol^-1. If enthalpy of vaporization of water is 10.5 k cal mol^-1 at 25° C, what would be enthalpy change for the hydration,

CuSO_4(s) +5H₂O_(l)→ CuSO₄.5H₂O_(s)

A
–57 kcal mol^–1

B
57 kcal mol^–1

C
19 kcal mol^–1

D
–19 kcal mol^–1

Solution

CuSO_4(s) + 5H₂O_(g)→ CuSO₄·5H₂O_(s),

ΔH = - 71.5 k cal

5H₂O_(e)→ 5H₂O(g), ΔH = 5 × 10.5 k cal

Adding the two eqn.,

CuSO_4(s) + 5 H₂O_(e)→ CuSO₄·5H₂O_(s)

ΔH = - 71.5 + 52.5 = - 19.0 k cal mol^-1
Question 4
1 / -0

A gas absorbs 200 J heat and undergoes simultaneous expansion against a constant external pressure of 10⁵ Pa. The volume changes from 4L to 5L. Find the change in internal energy.

A
–200 J

B
–100 J

C
200 J

D
100 J

Solution

ΔU = q + w = q – PΔV = 200J – 10⁵ P_a × (5 – 4) × 10^–3 m³
= 200J – 100 P_a m³ = 200 J – 100 J = 100 J
Question 5
1 / -0

Statement-1: At constant temperature and pressure whatever heat absorbed by the system is used in doing work.

Statement-2: Internal energy change is zero.

A
Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1.

B
Both Statement-1 and Statement-2 are true and Statement-2 is not the correct explanation of Statement-1.

C
Statement-1 is true but Statement-2 is false.

D
Statement-1 is false but Statement-2 is true.

Solution

Q = - W if ΔE = 0
Question 6
1 / -0

The heat of sublimation of iodine is 24 cal g^–1 at 50°C. If specific heat of solid iodine and its vapour are 0.055 and 0.031 cal g^–1 respectively, find the heat of sublimation of iodine at 100°C.

A
–22.8 cal g^–1

B
28.6 calg^–1

C
22.8 cal g^–1

D
–28. 6 cal g^–1

Solution

ΔH₂ - ΔH₁ = ΔC_p ( T₂ - T₁

ΔH₂ - 24 = ( 0.031 - 0.055 ) ( 100 - 50 )

⇒ ΔH₂ = 22.8 calg^-1
Question 7
1 / -0

If ΔH and ΔE are the change in enthalpy and change in internal energy respectively for a gaseous reaction, then

A
ΔH < ΔE, always

B
ΔH > ΔE, always

C
ΔH < ΔE, only if n_product> n_reactants

D
ΔH < ΔE, only if n_product< n_reactants

Solution

If n_p< n_r then Δn_(g) = n_p - n_r = '-'ve

Since ΔH = ΔE + ΔnRT

∴ ΔH = ΔE - some quantity [ D nis -ve ]

or, ΔH < ΔE

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Thermodynamics Test - 2

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Thermodynamics Test - 2

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Question 1

Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1.

Both Statement-1 and Statement-2 are true and Statement-2 is not the correct explanation of Statement-1.

Statement-1 is true but Statement-2 is false.

Statement-1 is false but Statement-2 is true.

Solution

Question 2

work done by the gas = 5 R

work done = 0

work done by the gas = 10 R

work done over the gas = 10 R

Solution

Question 3

–57 kcal mol–1

57 kcal mol–1

19 kcal mol–1

–19 kcal mol–1

Solution

Question 4

–200 J

–100 J

200 J

100 J

Solution

Question 5

Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1.

Both Statement-1 and Statement-2 are true and Statement-2 is not the correct explanation of Statement-1.

Statement-1 is true but Statement-2 is false.

Statement-1 is false but Statement-2 is true.

Solution

Question 6

–22.8 cal g–1

28.6 calg–1

22.8 cal g–1

–28. 6 cal g–1

Solution

Question 7

ΔH < ΔE, always

ΔH > ΔE, always

ΔH < ΔE, only if nproduct > nreactants

ΔH < ΔE, only if nproduct < nreactants

Solution

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–57 kcal mol^–1

57 kcal mol^–1

19 kcal mol^–1

–19 kcal mol^–1

–22.8 cal g^–1

28.6 calg^–1

22.8 cal g^–1

–28. 6 cal g^–1

ΔH < ΔE, only if n_product> n_reactants

ΔH < ΔE, only if n_product< n_reactants