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5 moles of neon gas at 2 atm and 27^{\circ} \mathrm{C} is adiabatically compressed to 1 / 3^{\text {rd }} its initial volume. \mathrm{\left[\gamma=1.67, \mathrm{C}_v=0.148 \mathrm{cal} / \mathrm{g}, \mathrm{M}=20.18 \mathrm{gm}\right. mole ]}
 

Option: 1

Pressure will change to \mathrm{6 \: atm.}


 


Option: 2

Temperature will become less than 600 \mathrm{~K}.
 


Option: 3

Work done by the gas is non-zero.
 


Option: 4

Work done on the gas in non-zero.


Answers (1)

best_answer

Using \mathrm{\mathrm{PV} \gamma= constant}

\mathrm{ P_f=P_1\left(\frac{V_1}{V_2}\right)=2 \times(3)^{1.67} \text { atm } \neq 6 \mathrm{~atm} }

Using \mathrm{TV { }^{\gamma-1}\: we \: get \: \mathrm{T}_2=\mathrm{T}_1\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}=300(3)^{0.67}}

Work is done on the gas since the volume change is non-zero and also the process is adiabatic.
 

Posted by

Gaurav

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