First Law of Thermodynamics - Free MCQ Practice Test with solutions, NEET

The IUPAC convention is as follow:-
For expansion or work done by the system on surrounding, W = negative
For compression or work done on system by surrounding, W = positive
Heat absorbed by the system is positive and heat released by the system is negative.
So, we have q = -65 J and W = 22 J
From 1st law of thermodynamics,
      ∆U = q+W
On putting values, we get ∆U = -43 J
So, q = -65 J, W = 22 J and ∆U = -43 J

The amount of mass or energy entering a control volume does not have to be equal to the amount of mass or energy leaving during an unsteady-flow process.It is mostly converted to internal energy as shown by a rise in the fluid temperature.

2CO + O2 ------------> 2CO₂
Atomic wt of C =12 kg or gms, Atomic wt of O = 16 kg or gms
2(12 + 16) kg of CO + 32 kg of O2 -----------> 2(12+32) kg of CO₂
56 kg of CO + 32 kg of O₂ -------> 88 kg of CO₂
For, 1kg of CO + 32/56 kg of O₂ ---------> 88/56 kg of CO₂
Therefore, 1kg of CO requires 32/56 or 4/7 kg of oxygen to produce 88/56kg or 11/7 kg of CO2

The expression for the change in internal energy is given by,
ΔU=q+w
q = heat absorbed = 200 J
w = work done = −P x V
=−2×10⁵ × 500×10⁻⁶
= −100 N - m
ΔU=200−100J= 100 J
Hence, the correct option is C.

When an ideal gas is compressed from the same initial to the same final state, the work done is given by

​

In a reversible compression, the external pressure is always equal to the gas pressure (P_ext = P_gas), so the pressure remains maximum at every stage. Hence, the area under the P ⁣− ⁣V curve (work) is maximum.

In an irreversible compression, the gas is compressed against a finite external pressure, so the pressure is lower during most of the process and the area under the curve is smaller.

Therefore, for compression between the same states,
W(I) > W(II).

As q = 0, we have adiabatic process.
V₁ = 100L and V₂ = 800L
T₁ = 300K and T₂ = not given
For NH₃, γ = 4/3
Applying TV^γ = constant
(300)(100)^4/3-1 = (T)(800)^4/3-1
T = 300/2 = 150K
W= nR(T₂-T₁)/γ-1
= 1×8.314×150/(4/3-1)
= 3714 J
= 900 cal

Thus,  T₂ =  150K and W = 900 cal
Thus option A is correct

Work done in isothermal process
W= -2.303 nRT log P1/P2
here n=1 and R =8.314  and T= 300 P1=4and P2=2.
W= -1729 J 
So option A is correct.

We know that power is energy per second. So by this definition, 187W means 187J/s energy is given. 
And 35J work is done.
So energy dissipated in the form of friction= 187 - 35 = 152J

Given:
Heat absorbed by the gas, Q = +50 J
Work done by the gas on surroundings, W = −100 J
First law of thermodynamics:
ΔU = Q − W
ΔU = 50 − (−100) = −50 J
For 1 mole of a diatomic gas:
Cv = (5/2)R ≈ 20.8 J mol^-1 K^-1
Change in temperature:
ΔT = ΔU / (nCv) = −50 / (1 × 20.8) ≈ −2.41 K
Negative sign indicates cooling.

Correct answer:
A: the gas will cool by 2.41°

T₁ = T₂ because process is isothermal
Work done in adiabatic process is less than in by isothermal process as area under isothermal curve is more than adiabatic curve.
In adiabatic processes, expansion is done by using internal energy. Hence it decreases while in the isothermal process, temperature remains constant. So   there is no change in internal energy.

Work done in isothermal process, w = nRTln(V₂/V₁)
w₆₀₀/w₃₀₀ = 1×R×600×ln(10)/ 1×R×300ln(10) = 2
∆U = 0 for isothermal processes.

Since the vessel is thermally insulated, q = 0
Further since,P_ext = 0, so w = 0,hence ∆U = 0
Since ∆T = 0, T,2 = T₁ and P₂V₂ = P₁V₁
However, the process is adiabatic irreversible, so we can’t apply P₂V₂^γ = P₁V₁^γ

1st law of thermodynamics states that:

δQ = ΔU + w

It is given that:

δQ = 254J

w = −73J because 73J of work is done on the gas

i.e. ΔU = 254 + 73 = 327J

From K to L; volume increases(sample is expanding) t = heating
From L to M; pressure decreases(molecules are moving far) = cooling
From M to N; volume decreases = cooling
From N to K; pressure increasing = heating

Isochoric process is that process where volume remains constant.
From the graph, we va say that for L→M and for N→K, volume remains constant.

Given:
Initial state:
P₁ = 10 bar, V₁ = 1 m³, T = 273 K
Final state:
P₂ = 1 bar, V₂ = 10 m³, T = 273 K
(Temperature is same, so overall process is isothermal.)
1 bar = 10⁵ Pa
I. Work done during expansion against constant external pressure (W₁)
External pressure = 1 bar = 10⁵ Pa
W₁ = Pext (V₂ − V₁)
W₁ = 10⁵ × (10 − 1)
W₁ = 9.0 × 10⁵ J
II. Work done on the system during compression (W₂)
External pressure = 10 bar = 10⁶ Pa
W₂ = Pext (V₁ − V₂)
W₂ = 10⁶ × (1 − 10)
W₂ = −9.0 × 10⁶ J
Magnitude of work done on the system = 9.0 × 10⁶ J

III. Total work done (W₃)
W₃ = W₁ + W₂
W₃ = 9.0 × 10⁵ − 9.0 × 10⁶
W₃ = −8.1 × 10⁶ J

IV. Least work required to restore original p–V conditions (W₄)
Least work occurs for reversible isothermal process:
W₄ = nRT ln(V₂ / V₁)
nRT = PV = 10 bar × 1 m³ = 10⁶ J
W₄ = 10⁶ ln(10)
W₄ ≈ 2.3 × 10⁶ J

Matching:
W₁ → 9.0 × 10⁵ J (q)
W₂ → 9.0 × 10⁶ J (q)
W₃ → 8.1 × 10⁶ J (p)
W₄ → 2.3 × 10⁶ J (r)

Blood → Gastric Juice
pH = 7.6021
[H+] = 2.5×10^-8 
∆G° = -RTlnK
= - RTln [1/2.5×10^-8]      (since, concentration of gastric juice be taken as unity and we have blood in reactant)
= 47443.26
Now for 1 mole(22.4L) we have Energy = 47443.26
So, for 1dm³(1 L), we have
   =47443.26/22.4
 =2 kJ
the correct answer is 2 

Cp - Cv = R
4R - Cv = R
Cv = 3R
γ = Cp/Cv = 4/3
T₁/T₂ = (v2/v1)^{(4/3 - 1)}
300/150 = (v2/v1)^1/3
2 = (v2/v1)^1/3
8 = v2/v1
v2 = 8v1
Therefore n = 8