Question

One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is $P_0$. Choose the correct option(s) from the following.

• A. Internal energies at A and B are the same
• B. Work done by the gas in process AB is $P_o{V}_o$ ln $4$
• C. Pressure at C is $\frac{P_o}{4}$
• D. Temperature at C is $\frac{T_o}{4}$

Answer 1

Answer: (A), (B)

The correct options are
A Internal energies at A and B are the same
B Work done by the gas in process AB is $P_o{V}_o$ ln $4$

From the graph given in the question, we can see that temperature at states Aand C are same,

i.e, $T_A=T_B$

Internal energy $=\frac{t}{2}nRT=\frac{t}{2}RT$ ( Here $n=1$ )

$\Rightarrow$ Here, $f=$ degree of freedom

$n=$ moles

$R=$ universal gas constant.

As, $f$ and $R$ are constant and $T_A=T_B,$

So, internal energy of the gas at A and B are same .

Hence, option A is correct.

For the process AB, temperature is constant throughout i.e, the process is isothermal.

Work done by a gas in an isothermal process $=nRT\iota \eta \left(\frac{v_2}{v_1}\right)$

$=\left(1\right)\left(R\right)\left(T_0\right)\iota \eta \left(\frac{v_B}{v_A}\right)$ [ Hence $n=1$ ]

$=R{T}_0\iota \eta \left(\frac{4v_0}{v_0}\right)$ [ On applying, $Pv=nRT$ at ${}^{\prime }{A}^{\prime },$ we will get $R{T}_0=p_0{v}_0$ ]

$=P_0{v}_0\iota \eta 4.$

So, option B is correct.

Hence, both option A and B are correct.