Two thermally insulated vessels 1 and 2 are filled with air at temperatures (T_{1},T_{2}), volume (V_{1},V_{2}), and pressure (P_{1},P_{2}), respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

  • Option 1)

    T_{1}+T_{2}\; \;

  • Option 2)

    (T_{1}+T_{2})/2\; \;

  • Option 3)

    \frac{T_{1}T_{2}(P_{1}V_{1}+P_{2}V_{2})}{P_{1}V_{1}T_{2}+P_{2}V_{2}T_{1}}\; \;

  • Option 4)

    \; \frac{T_{1}T_{2}(P_{1}V_{1}+P_{2}V_{2})}{P_{1}V_{1}T_{1}+P_{2}V_{2}T_{2}}

 

Answers (1)

As we learnt in 

Total internal energy -

U= U_{K}+U_{P}
 

- wherein

Change in internal energy

\Delta U= n\frac{f}{2}R\Delta T

(Always)

f is degree of freedom

 

 Let equilibrium temperature is T

Ui = Uf

n _{1}C_{v1} T_{1}+n_{2}C_{v2}T_{2}=\left ( \right n_{1}+n_{2})C_{v}T_{F}

C_{v1}=C_{v2}=C_{v}

\therefore T_{f}=\frac{n_{1}T_{2}+n_{2}T_{2}}{n_{1}+n_{2}}

n_{1}=\frac{P_{1}V_{1}}{RT_{1}},n_{2}=\frac{P_{2}V_{2}}{RT_{2}}

T_{f}=\frac{P_{1}V_{1}+P_{2}V_{2}}{\frac{P_{1}V_{1}}{T_{1}}+\frac{P_{2}V_{2}}{T_{2}}}

=\frac{\left (P_{1}V_{1}+P_{2}V_{2} \right )T_{1}T_{2}}{P_{1}V_{1}T_{2}+P_{2}V_{2}T_{1}}

 

 


Option 1)

T_{1}+T_{2}\; \;

This option is incorrect.

Option 2)

(T_{1}+T_{2})/2\; \;

This option is incorrect.

Option 3)

\frac{T_{1}T_{2}(P_{1}V_{1}+P_{2}V_{2})}{P_{1}V_{1}T_{2}+P_{2}V_{2}T_{1}}\; \;

This option is correct.

Option 4)

\; \frac{T_{1}T_{2}(P_{1}V_{1}+P_{2}V_{2})}{P_{1}V_{1}T_{1}+P_{2}V_{2}T_{2}}

This option is incorrect.

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