One mole of an ideal monoatomic gas is initially at 500K. If 300J of heat is added at constant volume, the final temperature of the gas is
(Assume R=8J/mol K)
A
525K
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B
550K
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C
600K
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D
625K
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Solution
The correct option is A525K Given, that gas is monoatomic
No. of moles of the gas (n)=1
Heat supplied (Q)=200J
Initial temperature (Ti)=500K
For monoatomic gas, CV=3R2
Since volume is constant, work done is zero.
From First law of thermodynamics, we know that ΔQ=ΔU+W W=0⇒ΔQ=ΔU ⇒ΔQ=ΔU=nCVΔT ⇒ΔQ=3R2×(Tf−Ti) ⇒300=32×8(Tf−500) ⇒Tf=525K is the final temperature.