The work done in an adiabatic change in a gas depends only on
Change is pressure
Change is volume
Change in temperature
None of the above
The pressure in the tyre of a car is four times the atmospheric pressure at $300 K$. If this tyre suddenly bursts, its new temperature will be $(\gamma = 1.4)$
In the following $P-V$ diagram two adiabatics cut two isothermals at temperatures $T_1$ and $T_2$ (fig.). The value of $\frac{{{V_a}}}{{{V_d}}}$ will be
The $PV$ diagram shows four different possible reversible processes performed on a monatomic ideal gas. Process $A$ is isobaric (constant pressure). Process $B$ is isothermal (constant temperature). Process $C$ is adiabatic. Process $D$ is isochoric (constant volume). For which process(es) does the temperature of the gas decrease ?
Two identical samples of a gas are allowed to expand $(i)$ isothermally $(ii)$ adiabatically. Work done is
Two different adiabatic paths for the same gas intersect two isothermal curves as shown in$P-V$ diagram. The relation between the ratio $\frac{V_a}{V_d}$ and the ratio $\frac{V_b}{V_c}$ is: