Question

A heat engine carries one mole of an ideal mono-atomic gas around the cycle as shown in figure, the amount of heat added in the process $AB$ and heat removed in the process $CA$ are:

• A. $q_{AB}=450R$ and $q_{CA}=-450R$
• B. $q_{AB}=450R$ and $q_{CA}=-225R$
• C. $q_{AB}=450R$ and $q_{CA}=-375R$
• D. $q_{AB}=375R$ and $q_{CA}=-450R$

Answer 1

The correct option is C $q_{AB}=450R$ and $q_{CA}=-375R$

Change in internal energy $\left(\Delta U\right)=n{C}_V\Delta T$

where,

n = no. of moles

$C_V=$ heat capacity at constant volume

$\Delta T=$ change in temperature

According to first law of thermodynamics,

$\Delta U=q+W$

$\Rightarrow q=\Delta U-W$

where,

$\Delta U=$ change in internal energy

$q$= heat added to system

$W=$ Work done by the system $=P\Delta V$

Case I:- For process AB,

$\Delta T=T_2-T_1=600-300=300K$

$n=1$

$C_V=\frac{3R}{2}\{\because \text{the gas is monoatomic}\}$

As V is constant,

$\Rightarrow \Delta V=0$

$\therefore W=0$

$\therefore q_{AB}=\Delta U+0=\Delta U=n{C}_V\Delta T=1\times \frac{3R}{2}\times 300=450R$

Case II:- For process CA,

$\Delta T=T_2-T_1=300-450=-150K$

$n=1$

$C_P=\frac{5R}{2}\{\because \text{the gas is monoatomic}\}$

$\therefore q_{CA}=n{C}_P\Delta T=1\times \frac{5R}{2}\times (-150)=-375R$