NEET-XII-Physics
24: Kinetic Theory of Gases
- #10The pressure of a gas kept in an isothermal container is 200 kPa. If half the gas is removed from it, the pressure will be
(a) 100 kPa
(b) 200 kPa
(c) 400 kPa
(d) 800 kPa.digAnsr: aAns : Let the number of moles in the gas be n.
Applying equation of state, we get
`` PV=nRT``
`` \Rightarrow P=\frac{nRT}{V}``
`` \Rightarrow 2\times {10}^{5}=\frac{nRT}{V}...\left(1\right)``
`` \,\mathrm{\,When\,}\,\mathrm{\,half\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,gas\,}\,\mathrm{\,is\,}\,\mathrm{\,removed\,},\,\mathrm{\,number\,}\,\mathrm{\,of\,}\,\mathrm{\,moles\,}\,\mathrm{\,left\,}\,\mathrm{\,behind\,}=\frac{\,\mathrm{\,n\,}}{2}``
`` \,\mathrm{\,Suppose\,}\,\mathrm{\,the\,}\,\mathrm{\,pressure\,}\,\mathrm{\,be\,}\,\mathrm{\,P\,}\text{'}.``
`` \,\mathrm{\,P\,}\text{'}=\frac{\,\mathrm{\,n\,}}{2}\frac{\,\mathrm{\,R\,}\,\mathrm{\,T\,}}{\,\mathrm{\,V\,}}``
`` \,\mathrm{\,Now\,},``
`` \,\mathrm{\,P\,}\text{'}=\frac{1}{2}\times 2\times {10}^{5}={10}^{5}\left[\,\mathrm{\,From\,}\,\mathrm{\,eq\,}.\left(1\right)\right]``
`` ``
`` ``
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=100 kPa
Thus,
(a) is the correct answer.