Question

A mono atomic ideal gas sample is given heat Q. One half of this heat is used as work done by the gas and rest is used for increasing its internal energy. The equation of process in terms of volume and temperature is ?



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Answer 1

Given:

A mono atomic ideal gas sample is given heat Q. One half of this heat is used as work done by the gas and rest is used for increasing its internal energy.

To find:

The equation of process in terms of volume and temperature .

Calculation:

Since half of the heat supply was used as work done and the rest was used for increasing the internal energy we can say the following statement:

$\boxed{ \sf{\Delta U =W = \dfrac{1}{2}Q}}$

Taking internal energy and work done in consideration :

$\sf{ \therefore \: \Delta U =W }$

$\sf{ = > \:n C_{v} dT=P \: dV }$

$\sf{ = > \:n C_{v} dT= \dfrac{nRT}{V} \times \: dV }$

For mono-atomic gas , C_(v) is 3R/2 ;

$\sf{ = > \:n \times \dfrac{3R}{2} \times dT= \dfrac{nRT}{V} \times \: dV }$

$\sf{ = > \dfrac{dV}{V} = \dfrac{3}{2} \times \dfrac{dT}{T} }$

Integrating on both sides:

$= > \displaystyle \int \sf{\dfrac{dV}{V}} = \dfrac{3}{2} \times \int \sf{ \dfrac{dT}{T} }$

$\sf{ = > ln(V) = \dfrac{3}{2} ln(T) }$

$\sf{ = > ln(V) = ln( {T}^{ \frac{3}{2} } ) }$

$\sf{ = > V = {T}^{ \frac{3}{2} } }$

$\sf{ = > {V}^{2} = {T}^{3}}$

So , final answer is :

$\boxed{ \red{ \huge{ \rm{ {V}^{2} = {T}^{3}}}}}$