Find net heat exchange for the process BC , if V B V C = 1 4 :

Process BC is isothermal W BC = − nRT ln V B V C = − 1 × R × 600 ln 1 4 = − 1200 R ln 2 In isothermal process Δ U = nC V Δ T ⇒ Δ U = 0 According to FLOT : Δ U = Q + W ⇒ 0 = Q + W ⇒ Q = − W ⇒ Q = + 1200 R ln 2

Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IIsothermal Process/Isobaric/IsochoricMedium2 minQB

ChemistryMediumsingle choice

Passage / Comprehension

Above graph is plotted for 1 mol of ideal monoatomic gas. Answer the following questions on the basis of above graph :

Find net heat exchange for the process BC , if $\frac{V _{C}}{V _{B}} = \frac{4}{1}$ :

Options:

Answer:

Solution:

Process BC is isothermal

$W_{BC} = - nRT ln \frac{V _{C}}{V _{B}} = - 1 \times R \times 600 ln \frac{4}{1} = - 1200 R ln 2$

In isothermal process

$Δ U = nC_{V} Δ T \Rightarrow Δ U = 0$

According to FLOT :

$Δ U = Q + W \Rightarrow 0 = Q + W \Rightarrow Q = - W$

$\Rightarrow Q = + 1200 R ln 2$

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

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...