Question

An ideal gas ( $1$ mole, monatomic) is in the initial state $P$ (see diagram) on an isothermal curve $A$ at a temperature $T_0$ . It is brought under a constant volume $\left(2V_0\right)$ process to $Q$ which lies on an adiabatic curve $B$ intersecting the isothermal curve $A$ at $\left(P_0,V_0,T_0\right)$ . The change in the internal energy of the gas (in terms of $T_0$ ) during the process is $\left(2^{2/3}=\text{1.587}\right)$

• A. $\text{2.3}{T}_0$
• B. $\text{-4.6}{T}_0$
• C. $\text{-2.3}{T}_0$
• D. $\text{4.6}{T}_0$

Answer 1

Answer: (B)

Temperature at state $P=T_0$ , since $P$ lies on the isotherm of temperature $T_0$ . If $T$ be the temperature at $Q$ , then for the adiabatic process $B$ , we have $T_0{V}_0^{\gamma -1}=T{\left(2V_0\right)}^{\gamma -1}$
$T=\frac{T}{2^{\gamma -1}}=\frac{T_0}{2^{2/3}}$
Change in the internal energy of the gas is
$\Delta U=C_V\left(T-T_0\right)=\left(\frac{R}{-1}\right)\left(\frac{T_0}{2^{2/3}}-T_0\right)$
$=\frac{3R{T}_0\left(1-2^{2/3}\right)}{2\times 2^{2/3}}=-\text{4.6}{T}_0$