An ideal gas, undergoing adiabatic change, has which of the following pressure temperature relationship? (1996)
From the adiabatic relation $PV^\gamma = \text{constant}$ and the ideal gas law $PV = nRT$, eliminating volume yields $P^{1-\gamma} T^\gamma = \text{constant}$.
A diatomic gas initially at $18^\circ\text{C}$ is compressed adiabatically to one eighth of its original volume. The temperature after compression will be: (1996)
Using the relation $T V^{\gamma-1} = \text{constant}$ with $\gamma = 1.4$ for a diatomic gas and $V_2 = V_1 / 8$, we find $T_2 = 291 \times (8)^{0.4} \approx 668.3\text{ K}$, which is $395.3^\circ\text{C}$.
In an adiabatic change, the pressure and temperature of a monoatomic gas are related as $P \propto T^C$ where $C$ equals: (1994)
From $P^{1-\gamma} T^\gamma = \text{constant}$, we get $P \propto T^{\frac{\gamma}{\gamma-1}}$. For a monoatomic gas, $\gamma = 5/3$, so $C = \frac{5/3}{5/3 - 1} = \frac{5}{2}$.
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 $P_3 = (P/4)(4)^{1.5} = P \times 4^{0.5} = 2P$.
The efficiency of a Carnot engine operating with reservoir temperature of $100\text{ }^\circ\text{C}$ and $-23\text{ }^\circ\text{C}$ will be: (1997)
Efficiency $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{250\text{ K}}{373\text{ K}} = \frac{373-250}{373}$.
An ideal Carnot engine, whose efficiency is $40\%$, receives heat at $500\text{ K}$. If its efficiency is $50\%$, the intake temperature for the same exhaust temperature is: (1995)
Exhaust temperature $T_2 = 500(1-0.4) = 300\text{ K}$. New intake temperature $T_1' = \frac{300}{1-0.5} = 600\text{ K}$.