Sunil Kumar FP
Last Activity: 11 Years ago
work done can be calculated if we consider it as an isothermal reversible process
initial condition,PV=nRT,
1×5=1×.0821×T
T=60.9K
Now workdone in an isothermal reversible process= -2.303nRTlog V2/V1
-2.303×1×.0821×60.9log 50/5
=-140.25 joule(negative sign because work is done by the system)
thanks and regard
sunil kumar
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