AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

< div class="hscrollenable">< span class="sprite AP_23_Sep_2020_S1_112sol">< \/span>< \/div>Given that, pA=5kPa=5×103Pa
pB=2kPa=2×103Pa
VA‌‌=4m3
VB‌‌=6m3
µ‌‌=1‌ mole ‌
Change in internal energy, ∆U=µCV∆T
=µ(‌

5

2

R)(TB−TA)
( ∵ For diatomic gas, CV=‌

5

2

R)
∴‌‌∆U=‌

5

2

µR(TB−TA)
‌‌=‌

5

2

(µRTB−µRTA)
‌=‌

5

2

(pBVB−pAVA)‌‌‌ (using ‌pV=µRT)
‌‌=‌

5

2

(2×103×6−5×103×4)
‌‌=‌

5

2

×(−8×103)=−20×103=−20kJ