Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A monatomic ideal gas of two moles is taken through a cyclic process starting for $\mathrm{A}$ as shown in the figure. The volume ratios are $\mathrm{\left(V_B / V_A\right)=2 \: and \: \left(V_D / V_A\right)=4}$ . If the Temperature $\mathrm{T}_{\mathrm{A}}$ at $\mathrm{A}\: \text{is}\: 27^{\circ} \mathrm{C}$ . The total work done by the gas during the complete cyc Is:

Option: 1
$\mathrm{200\: R}$

Option: 2
$\mathrm{400\: R}$

Option: 3
$\mathrm{600\: R}$

Option: 4
$\mathrm{800\: R}$

Answers (1)

best_answer

Taking $\mathrm{V_A=V_0;}$ $\mathrm{ V_B=2 V_0 ; V_D=4 V_0 }$

Process $\mathrm{ A \rightarrow B }$ : Isobaric process

as $\mathrm{ \mathrm{V} / \mathrm{T}= }$ constant

$\mathrm{ \mathrm{V}_0 /(273+27)=\left(2 \mathrm{~V}_0\right) / \mathrm{T}_{\mathrm{B}} }$

$\mathrm{ \Rightarrow \mathrm{T}_{\mathrm{B}}=600 \mathrm{~K}=327^{\circ} \mathrm{C}. }$

$\mathrm{ Q_1=n C_P \Delta T=2(5 / 2) R(600-300) }$

$\mathrm{ =1500 \mathrm{R} (absorbed) }$

Process $\mathrm{ B \rightarrow C }$ (Isothermal)

$\mathrm{ Q_2=W_2=n R T_B \ln \left(V_2 / V_1\right) }$

$\mathrm{ =831.78 \mathrm{R} (absorbed) }$

Process $\mathrm{ C \rightarrow D }$ (Isochoric)

$\mathrm{ Q_3=n C_V \Delta T=-900 R }$ (released)

Process $\mathrm{ D \rightarrow A }$ ( Isothermal)

$\mathrm{ Q_4=W_4=n R T_A \ln \left(V_2 /V_1\right)=-831.78 R }$

Total work done: $\mathrm{ W=Q_1+Q_2+Q_3+Q_4=600 R. }$

Posted by

avinash.dongre

View full answer

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET UG Paper Leak 2026 LIVE: NTA response sheet soon at neet.nta.nic.in; RSOG investigation on paper leak

anam-khan

NEET UG 2026 irregularities under investigation; NTA says full cooperation being extended to agencies

anam-khan

NEET UG 2026 Tie-Breaking Rule Explained: Here’s how ranks are decided

Latest Question