2moles of an ideal diatomic gas is initially at 300K. If heat of 300J is added to it at constant pressure, then the final temperature of the gas is
A
30.5K
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B
305.15∘C
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C
305.15 K
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D
30.5∘C
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Solution
The correct option is C305.15 K Given, that the gas is diatomic ∴CV=5R2⇒CP=7R2 Heat supplied (Q)=300J From, Q=nCPΔT Q=n7R2×(Tf−Ti) 300=2×7×8.314×(Tf−300)2 ⇒Tf=305.15K