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Two moles of an ideal monoatomic gas occupies a volume V at 27oC. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

  • Option 1)

    (a) 195 K         (b) 2.7 kJ

  • Option 2)

    (a) 189 K         (b) 2.7 kJ
        

  • Option 3)

    (a) 195 K         (b) −2.7 kJ

  • Option 4)

    (a) 189 K         (b) −2.7 kJ

 

Answers (2)

best_answer

As we learnt that

 

Equation of state -

dQ= 0

n\, C_{V}\, dT+PdV= 0
 

- wherein

On solving

\gamma \frac{dV}{V}+\frac{dP}{P}= 0

\Rightarrow PV^{\gamma }= constant

 

 For Adibatic process

T_{1}v_{1}^{r-1}=T_{2}v_{2}^{r-1}          (r=5/3)

300(v)^{2/3}=T_{2}(2v)^{2/3}

or T_{2}=\frac{300}{2^{2/3}}\simeq 189K

\Delta v=\frac{f}{2}(nR\Delta T)=\frac{3}{2}.2.\frac{25}{3}.(189-300)5

       = -2.7kJ


Option 1)

(a) 195 K         (b) 2.7 kJ

This is incorrect

Option 2)

(a) 189 K         (b) 2.7 kJ
    

This is incorrect

Option 3)

(a) 195 K         (b) −2.7 kJ

This is incorrect

Option 4)

(a) 189 K         (b) −2.7 kJ

This is correct

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