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Question Number 42131 by Necxx last updated on 18/Aug/18
$${Determine}\:{the}\:{total}\:{translational} \\ $$$${energy}\:{of}\:{two}\:{litres}\:{of}\:{oxygen}\:{gas} \\ $$$${held}\:{at}\:{a}\:{temperature}\:{of}\:\mathrm{0}°{C}\:{and}\: \\ $$$${pressure}\:{of}\:{one}\:{atmosphere}. \\ $$$$ \\ $$
Commented by Necxx last updated on 19/Aug/18
$${please}\:{help} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Aug/18
$${PV}={nRT}\:\:\:\:\:\:{P}={pressure}\:\:\:\:{V}={volume}\:\:\:\:{T}={temp} \\ $$$${n}=\frac{{PV}}{{RT}}\:\:\:{n}={numbers}\:{of}\:{mole} \\ $$$${U}=\frac{\mathrm{1}}{\mathrm{2}}{NfKT}\:\:\:\:{U}={total}\:{energy}\:\:{N}={avogadro}\:{number} \\ $$$${f}={degrees}\:{of}\:{freedom}\:\:\:{NK}={R} \\ $$$$\:\:\:\:\:{U}=\frac{\mathrm{1}}{\mathrm{2}}{RfT}\:\:\:\: \\ $$$${for}\:{oxygen}\:{f}=\mathrm{3}\:{translational}\:{degrees}\:{offrredom} \\ $$$${energy}\:{per}\:{degrees}\:{of}\:{freedom} \\ $$$$\frac{{U}}{{f}}=\frac{\mathrm{1}}{\mathrm{2}}{RT} \\ $$$${so}\:{required}\:{translational}\:{energy} \\ $$$$=\mathrm{3}{n}×\frac{\mathrm{1}}{\mathrm{2}}×{RT} \\ $$$$=\mathrm{3}×\frac{{PV}}{{RT}}×\frac{\mathrm{1}}{\mathrm{2}}×{RT} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\left({PV}\right)=\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{1}.\mathrm{01}×\mathrm{10}^{\mathrm{5}} ×\mathrm{2}×\mathrm{10}^{−\mathrm{3}} =\mathrm{3}.\mathrm{03}×\mathrm{10}^{\mathrm{2}} {J} \\ $$$${pls}\:{check}.... \\ $$$$ \\ $$
Commented by Necxx last updated on 20/Aug/18
$${oh}...{Thanks}\:{boss} \\ $$