Gaseous hydrogen contained initially under standard conditions in a sealed vessel of volume V=5L was cooled by ΔT=55K. The internal energy of the gas will change by
A
−255J
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B
200J
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C
250J
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D
−200J
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Solution
The correct option is A−255J
Volume of vessel V=5L=0.005m3
Standard conditions : P=1 atm =1.01325×105 Pa T=273.15K
Using Ideal gas equation : PV=nRT
∴(1.01325×105)×0.005=nR×273.15⟹nR=1.8547
Degree of freedom for diatomic gas f=5
Change in internal energy ΔU=f2nRΔT=52×1.8547×(−55)=−255J