Two identical balls, $A$ and $B$ , of uniform composition and initially at the same temperature, each absorb exactly the same amount of heat. $A$ is hanging down from the ceiling while $B$ rests on the horizontal floor in the same room. Assuming no subsequent heat loss by the balls, which of the following statements is correct about their final temperatures, $T_A$ and $T_B$ , once the balls have reached their final state?
$T_A < T_B$
$T_A > T_B$
$T_A = T_B$
$T_A \leq T_B$
$\Delta U + \Delta W = 0$ is valid for
A gas is being compressed adiabatically. The specific heat of the gas during compression is
Match the thermodynamic processes taking place in a system with the correct conditions. In the table: $\Delta Q$ is the heat supplied, $\Delta W$ is the work done and $\Delta U$ is change in internal energy of the system
Process | Condition |
$(I)$ Adiabatic | $(A)\; \Delta W =0$ |
$(II)$ Isothermal | $(B)\; \Delta Q=0$ |
$(III)$ Isochoric | $(C)\; \Delta U \neq 0, \Delta W \neq 0 \Delta Q \neq 0$ |
$(IV)$ Isobaric | $(D)\; \Delta U =0$ |
Consider one mole of helium gas enclosed in a container at initial pressure $P_1$ and volume $V_1$. It expands isothermally to volume $4 V_1$. After this, the gas expands adiabatically and its volume becomes $32 V_1$. The work done by the gas during isothermal and adiabatic expansion processes are $W_{\text {iso }}$ and $W_{\text {adia, }}$ respectively. If the ratio $\frac{W_{\text {iso }}}{W_{\text {adia }}}=f \ln 2$, then $f$ is. . . . . . . .
Starting at temperature $300\; \mathrm{K},$ one mole of an ideal diatomic gas $(\gamma=1.4)$ is first compressed adiabatically from volume $\mathrm{V}_{1}$ to $\mathrm{V}_{2}=\frac{\mathrm{V}_{1}}{16} .$ It is then allowed to expand isobarically to volume $2 \mathrm{V}_{2} \cdot$ If all the processes are the quasi-static then the final temperature of the gas (in $\left. \mathrm{K}\right)$ is (to the nearest integer)