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 $P_3 = (P/4)(4)^{1.5} = P \times 4^{0.5} = 2P$.
When volume changes from $V$ to $2V$ at constant pressure $P$, then the change in internal energy will be: (1998)
Work done $\Delta W = P(2V - V) = PV$. Heat supplied $\Delta Q = n C_p \Delta T = \frac{\gamma PV}{\gamma-1}$. Thus, change in internal energy $\Delta U = \Delta Q - \Delta W = \frac{PV}{\gamma-1}$.
A gas of volume changes $2\text{ litre}$ to $10\text{ litre}$ at constant temperature $300\text{ K}$, then the change in internal energy will be: (1998)
Since the process takes place at a constant temperature (isothermal), the internal energy of an ideal gas depends only on temperature, so the change in internal energy is zero.
A sample of gas expands from volume $V_1$ to $V_2$. The amount of work done by the gas is greatest, when the expansion is: (1997)
On a $P-V$ diagram, the work done is represented by the area under the curve. For the same expansion volume, isobaric expansion maintains the highest pressure throughout, resulting in the maximum area and work done.
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 $C = \frac{5/3}{5/3 - 1} = \frac{5}{2}$.