Kinetic Theory

JEE Main level, single correct, medium

13 of 168

Question 13 of 168

A vessel contains 16g16g of hydrogen and 128g128g of oxygen at standard temperature and pressure. The volume of the vessel in cm3{cm}^{3} is :

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Answer: (C) 27×10427\times {10}^{4}

Number of moles of Hydrogen (H2)=mM=162=8\left({H}_{2}\right)=\dfrac{m}{M}=\dfrac{16}{2}=8
Number of moles of oxygen (O2)=12832=4\left({O}_{2}\right)=\dfrac{128}{32}=4
Using ideal gas equation, PV=nRTPV=nRT,
Where pressure P=1atm=1.01×105Pa(STP)P=1atm=1.01\times {10}^{5}Pa\left(STP\right)
Putting the values,
1.01×105V=(8+4)×8.314×2731.01\times {10}^{5}V=\left(8+4\right)\times 8.314\times 273
V=12×8.314×2731.01×105m3V=\dfrac{12\times 8.314\times 273}{1.01\times {10}^{5}}{m}^{3}
=0.27×106cm3=27×104cm3=0.27\times {10}^{6}{cm}^{3}=27\times {10}^{4}{cm}^{3}

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