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Two moles of Helium gas \mathrm{(\gamma=5 / 3)} are initially at temperature \mathrm{27^{\circ} \mathrm{C}} and occupy a volume of 20 litres. The gas is first expanded at the constant pressure until volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value. What are the final volume.
 

Option: 1

\quad 113.13 \times 10^{-3} \mathrm{~m}^3
 


Option: 2

\quad 213.13 \times 10^{-3} \mathrm{~m}^3
 


Option: 3

313.13 \times 10^{-3} \mathrm{~m}^3
 


Option: 4

13.13 \times 10^{-3} \mathrm{~m}^3


Answers (1)

best_answer

From ideal gas equation

\mathrm{ \mathrm{PV}=\mathrm{nRT} }

\mathrm{ \text { initial pressure } \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}=\frac{2 \times 8.3 \times 300}{20 \times 10^{-3}} }

\mathrm{ =2.49 \times 10^5 \mathrm{~N} / \mathrm{m}^2}

When volume of gas is doubled at constant pressure, its temperature is also doubled. This process is shown on P-V curve by line AB. The gas then cools to temperature T adiabatically. This is shown by curve BC. The whole process is represented by curve \mathrm{ A B C.}

At point B, pressure \mathrm{P_B=P_A=2.49 \times 10^5 \mathrm{~N} / \mathrm{m}^2}.Volume \mathrm{V_B=2 V_A=40 \times 10^{-3} \mathrm{~m}^3,} Temperature \mathrm{\mathrm{T}_{\mathrm{B}}=600 \mathrm{~K}.}
Now from adiabatic equation \mathrm{\mathrm{TV}^{\gamma-1}=} constant

We have \mathrm{\mathrm{T}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}{ }^{(\gamma-1)}=\mathrm{T}_{\mathrm{C}} \mathrm{V}_{\mathrm{C}}{ }^{(\gamma-1)}}

\mathrm{\therefore \left(\frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{C}}}=\frac{600}{300}=2 }

\mathrm{\therefore \frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}}=2^{1 /(\gamma-1)}=2^{3 / 2}}

Final volume

\mathrm{ \therefore \quad V_C=2 \sqrt{2} V_B }

\mathrm{ =2 \times 1.414 \times 40 \times 10^{-3}=113.13 \times 10^{-3} \mathrm{~m}^3}





 

Posted by

shivangi.shekhar

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