At ${27^o}C$ a gas is suddenly compressed such that its pressure becomes $\frac{1}{8}th$ of original pressure. Temperature of the gas will be $(\gamma = 5/3)$
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$n$ moles of a van der Waals' gas obeying the equation of state $\left(p+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$, where $a$ and $b$ are gas dependent constants, is made to undergo a cyclic process that is depicted by a rectangle in the $p-V$ diagram as shown below. What is the heat absorbed by the gas in one cycle?
An ideal gas goes through a reversible cycle $a\to b\to c\to d$ has the $V - T$ diagram shown below. Process $d\to a$ and $b\to c$ are adiabatic.... The corresponding $P - V$ diagram for the process is (all figures are schematic and not drawn to scale)
A monoatomic ideal gas, initially at temperature ${T_1},$ is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature. ${T_2}$ by releasing the piston suddenly. If ${L_1}$ and ${L_2}$ are the lengths of the gas column before and after expansion respectively, then ${T_1}/{T_2}$ is given by
An ideal gas is taken reversibly around the cycle $a-b-c-d-a$ as shown on the temperature $T$ - entropy $S$ diagram. The most appropriate representation of above cycle on a internal energy $U$ - volume $V$ diagram is