Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IIsothermal Process/Isobaric/IsochoricMedium2 minQB

ChemistryMediummultiple choice

Passage / Comprehension

2 moles of an ideal gas are subjected to two different processes (separately).

Process A: $(2 ba r, 300 K)$ to $(2 ba r, 450 K)$ isobarically.

Process B: $(30 l i t re, 300 K)$ to $(30 l i t re, 450 K)$ isochorically.

Given: $C_{v, m} = \frac{3}{2} R$

$R = 2 c a l / m o l - K$

Which of the following is CORRECT for process B

Answer:

A, B, D

Solution:

For process B (isochoric process), volume is constant, so work done $W = 0$ .

Hence, by the First Law of Thermodynamics, $Q = Δ U$ .

$Δ T = 450 - 300 = 150 K$

$C_{V} = \frac{3}{2} R = 3 c a l / m o l - K$

$Δ U = n C_{V} Δ T = 2 \times 3 \times 150 = 900 c a l \approx 12 k J$

For enthalpy:

$C_{P} = C_{V} + R = 5 c a l / m o l - K$

$Δ H = n C_{P} Δ T = 2 \times 5 \times 150 = 1500 c a l \approx 21 k J$

In an isochoric process, pressure increases, so the $Δ (P V)$ change corresponds to $n R Δ T \approx 9 k J$ .

Hence, the correct options are **A, B and D**.

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Isothermal Process/Isobaric/Isochoric

⏱ 2mℹ️ Source: QB

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