Thermodynamics - Free MCQ Practice Test with solutions, NEET Chemistry

Work done =- ∫_Vi^Vf p_ext dVwhere V_i is initial and Vf is final volume. P_ex is the external pressure applied on system while change in volume

-ve sign is used as final volume v_f will be less than initial volume v_i  (v_f<v_i)
so to make final answer +ve a -ve sign is used .

H =U+pV. H is enthalpy of reaction.

Entropy is a state function and it is measure of randomness of a system.

Spontaneity means a reaction occurring on its own without any help of external agency.

when 1 mole of CO₂ is produced energy released is –393.5 kJ mol⁻¹ Moles of CO₂ given =35.2/44 =0.8 moles So energy released = 0.8 x393.5 KJ/mol = 315 KJ/mol

Specific heat will remain constant.

The rate H at which heat is transferred through the slab is,

(a) directly proportional to the area (A) available.

(b) inversely proportional to the thickness of the slab Δx.

(c) directly proportional to the temperature difference ΔT.

So, H = kA ΔT/ Δx    

Where k is the proportionality constant and is called thermal conductivity of the material.

From above we know that the rate H at which heat is transferred through the slab is directly proportional to the area (A) available.

Area A of the solid sphere is defined as,

A = 4πr²

Here r is the radius of the sphere.

So, the area A₁ of a uniform small solid sphere having radius r₁ will be,

A₁ = 4πr₁²

And, the area A₂ of a uniform large solid sphere having radius r₂ will be,

A₂ = 4πr₂²

Thus the area A from which heat is transferred through the surface of the sphere will be the difference of area of uniform large solid sphere A₂ and small solid sphere A₁.

So, A = A₂ - A₁ = 4πr₂² - 4πr₁² = 4π (r₂² - r₁²)

Since the rate H at which heat is transferred through the slab is directly proportional to the area (A) available, therefore the rate at which heat is transferred through the surface of the sphere is proportional to r₂² - r₁².

Q_reversible = T∆S
Hence, Q = To(S2-S1)

The molar heat capacity of a substance is the quantity of heat needed to raise the temperature of one mole by one degree Celsius.

ΔG < 0 means reaction is spontaneous. And ΔG >0 means reaction is non spontanenous.

C₄H₁₀(g) + 13/2O₂  →  4CO₂(g) + 5H₂O(l)
∆N_g = (n_g)_product - (n_g)_reactant
= 4-(13/2+1) = -7/2
or ∆N_g is negative.
∆H = ∆U + ∆N_gRT
SInce ∆n_g is negative therefore ∆H is less than ∆U.

for free expansion P_ext =0 so w=0 and for adiabatic process q=0 therefore ΔU =0 making ΔT=0

For ideal gas C_p - C_V = nR. For1 mole of gas n=1.

The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation is enthalpy of formation.

State function is one which is only dependent on initial and final state of the system and is independent of the path by which that change has occurred.

Thermodynamics is not concerned about how and at what rate these energy transformations are carried out, but is based on initial and final states of a system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state.

w (reversible) < w (irreversible) (for compression process)

• Work done is the area under the P−V curve. It can be seen in the curve above that the area under the P−V curve for irreversible compression of the gas is more than the area under the curve for reversible compression.
• Thus, work done for irreversible compression is more than that for reversible compression.

since volume is constant so w=0 so q=ΔE=nC_vΔT C_v=(3/2)R

Because  argon is monoatomic, approximately ideal gas,

c_v = 3/2 R = 12.47 J K^-1 mol^-1

At constant volume,

q_v = n c_v Δ T

1000J = (2.00 mol)(12.47 JK^-1mol^-1)ΔT

ΔT = 40.1 k T_f = 298 + 40.1 =338 K

Exothermic process results in decrease in enthalpy of system because for exothermic reaction ΔH= -ve.

For adiabatic process Δq=0.

For an isolated system, ΔU = 0. According to the second law of thermodynamics, any spontaneous process in an isolated system leads to an increase in entropy. For example, when two gases A and B mix freely in an isolated container, the system becomes more disordered and ΔS > 0 while ΔU = 0.

For freezing of process since process is spontaneous therefore if ΔS (system) decreases but ΔS (surroundings) increases.