Question

The PV-graph for a monatomic gas is shown in figure. Find the energy absorbed by the gas
during this process.
​

Answer 1

Given:

The PV-graph for a monatomic gas is shown in figure.

To find:

Heat absorbed during the process.

Calculation:

Internal Energy Change of the gas

$\sf{\therefore \: \Delta U = \mu \times C_{V} \times d \theta}$

$\sf{ = > \: \Delta U = \mu \times \dfrac{3R}{2} \times d \theta}$

$\sf{ = > \: \Delta U = \mu \times \dfrac{3R}{2} \times \bigg \{\dfrac{(3P_{0})(2V_{0})}{ \mu R } - \dfrac{(2P_{0})(V_{0})}{ \mu R } \bigg \}}$

$\sf{ = > \: \Delta U = \mu \times \dfrac{3R}{2} \times \bigg \{\dfrac{6P_{0}V_{0}}{ \mu R } - \dfrac{2P_{0}V_{0}}{ \mu R } \bigg \}}$

$\sf{ = > \: \Delta U = \mu \times \dfrac{3R}{2} \times \bigg \{\dfrac{4P_{0}V_{0}}{ \mu R } \bigg \}}$

$\sf{ = > \: \Delta U =6P_{0}V_{0}}$

Now, work done

$\sf{W = \dfrac{1}{2} \bigg( 2P_{0} + 3P_{0} \bigg ) \bigg(2V_{0} - V_{0} \bigg)}$

$= > \sf{W = \dfrac{1}{2} \bigg( 5P_{0} \bigg ) \bigg(V_{0} \bigg)}$

$= > \sf{W = \dfrac{ 5P_{0} V_{0}}{2} }$

So, heat change

$\sf{\Delta Q = \Delta U + W}$

$= > \sf{\Delta Q = 6P_{0}V_{0} + \dfrac{5P_{0}V_{0}}{2} }$

$= > \sf{\Delta Q = \dfrac{17P_{0}V_{0}}{2} }$

So, final answer is:

$\boxed{ \red{ \sf{\Delta Q = \dfrac{17P_{0}V_{0}}{2} }}}$