Question

a. On the axes below i. Draw the PV-Diagram representing the series of processes; 6T 0 0 0 2V 3%

Determine the heat transferred during process 2 (from 2T_o to 6T_o) in terms of P_o and V_o.

Answer 1

Work Done during the process 2 is $\Delta W= -p\Delta V= -2p_0(3V_0-V_0)= -4p_0V_0$

Internal Energy Change during the process 2 is

   $\Delta U= nc_V\Delta T= nc_V(\frac{P\Delta V}{nR})$   ( Using Ideal Gas Equation and c_V= heat capacity at constant volume)

Now using the given values,

$\Delta U= c_V(\frac{2p_0(3V_0-V_0)}{R})$

$\Delta U= c_V(\frac{4p_0V_0}{R})$

Then using 1st law of Thermodynamics,

   $\Delta Q= \Delta U-\Delta W$

Using above calculated values,

$\Delta Q= c_V(\frac{4p_0V_0}{R})+4p_0V_0$

$\Delta Q= 4p_0V_0(\frac{c_v}{R} + 1)$

$\Delta Q= 4p_0V_0(\frac{c_v+ R}{R})= 4p_0V_0(\frac{c_p}{R})$ ( Using MAYER'S relation)

( where c_p= specific heat capacity at constant pressure)