Determine the temperature (in Kelvin) for 1 mol of gas at 10 atm with a volume of 15L
Answer
No. Of moles n= 1
Volume
V=15L=15×10−3m315\times10^{-3}m^315×10−3m3
Presure
P=1atm=1.01×105Pa1.01\times10^{5}Pa1.01×105Pa
Using ideal gas equation
PV=nRTPV=nRTPV=nRT
So temperature
T=PVnRT=\frac{PV}{nR}T=nRPV
Putting all values
T=1.01×105×15×10−31×8.314=182KT=\frac{1.01\times10^5\times15\times10^{-3}}{1\times8.314}\\=182KT=1×8.3141.01×105×15×10−3=182K
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