In which of the following processes, heat is neither absorbed nor released by a system? (2019)
In an adiabatic process, the system is insulated from its surroundings, meaning the net heat exchange ($Q$) is zero.
A sample of $0.1 \text{ g}$ of water at $100^{\circ}\text{C}$ and normal pressure ($1.013 \times 10^{5} \text{ Nm}^{-2}$) requires $54 \text{ cal}$ of heat energy to convert to steam at $100^{\circ}\text{C}$. If the volume of the steam produced is $167.1 \text{ cc}$, the change in internal energy of the sample, is: (2018)
Heat supplied $Q = 54 \text{ cal} = 54 \times 4.18 \text{ J} = 225.72 \text{ J}$. Work done $W = P\Delta V = 1.013 \times 10^{5} \times (167.1 - 0.1) \times 10^{-6} \approx 16.92 \text{ J}$. By first law of thermodynamics, $\Delta U = Q - W = 225.72 - 16.92 = 208.8 \text{ J}$, which is closest to $208.7 \text{ J}$.
14. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of $\frac{C_p}{C_v}$ for the gas is: (2013)
$P \propto T^3$ or $P T^{-3} = \text{constant}$. We know that for an adiabatic process $P^{1-\gamma} T^{\gamma} = \text{constant}$ or $P T^{\frac{\gamma}{1-\gamma}} = \text{constant}$. Comparing the powers of $T$, $\frac{\gamma}{1-\gamma} = -3 \Rightarrow \gamma = -3 + 3\gamma \Rightarrow 2\gamma = 3 \Rightarrow \gamma = 3/2$.
11. An ideal gas is compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas? (2015 Re)
Work done on the gas is maximum in the adiabatic process as the area under the $P-V$ curve is maximum.
13. A monoatomic gas at a pressure $P$, having a volume $V$ expands isothermally to a volume $2V$ and then adiabatically to a volume $16V$. The final pressure of the gas is (take $\gamma = 5/3$): (2014)
For isothermal process $P_1V_1 = P_2V_2 \Rightarrow P \times V = P_2 \times 2V \Rightarrow P_2 = P/2$. For adiabatic process $P_2V_2^{\gamma} = P_3V_3^{\gamma} \Rightarrow (P/2)(2V)^{5/3} = P_3(16V)^{5/3} \Rightarrow P_3 = P/64$.
In thermodynamic processes which of the following statements is not true? (2009)
An isochoric process is one in which volume remains constant, not pressure. Therefore, statement (a) is incorrect.
The internal energy change in a system that has absorbed $2\text{ kcal}$ of heat and done $500\text{ J}$ of work is: (2009)
Using $\Delta Q = 2\text{ kcal} = 2000 \times 4.2\text{ J} = 8400\text{ J}$ and $\Delta W = 500\text{ J}$. From the first law, $\Delta U = \Delta Q - \Delta W = 8400 - 500 = 7900\text{ J}$.
If $Q$, $E$ and $W$ denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then: (2004)
In a closed cycle process, the system returns to its initial state, meaning there is no net change in internal energy ($E = 0$ or $\Delta U = 0$).
One mole of an ideal gas at an initial temperatures of $T\text{ K}$ does $6\text{ R}$ joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $\frac{5}{3}$, the final temperatures of gas will be: (2004)
In an adiabatic process, $\Delta U = -W = -6R$. Also, $\Delta U = n C_v \Delta T = \frac{R}{\gamma - 1}(T_2 - T)$. With $\gamma = 5/3$, we get $\frac{3}{2}(T_2 - T) = -6$, leading to $T_2 = T - 4\text{ K}$.
Initial pressure and volume of a gas are $P$ and $V$ respectively. First its volume is expanded to $4V$ by isothermal process and then again its volume makes to be $V$ by adiabatic process, then its final pressure is ($\gamma = 1.5$): (1999)
After isothermal expansion, pressure becomes $P/4$ at volume $4V$. Following adiabatic compression back to volume $V$, the final pressure is $P_3 = (P/4)(4)^{1.5} = P \times 4^{0.5} = 2P$.