Helium at $27^oC$ has a volume of $8$ litres. It is suddenly compressed to a volume of $1$ litre. The temperature of the gas will be ....... $^oC$ $[\gamma = 5/3]$

$5.6\, L$ of helium gas at $STP$ is adiabatically compressed to $0.7\, L$. Taking the initial temperature to be $T_1$,the magnitude of work done in the process is

$A$ certain volume of a gas at $300 \ K$ expands adiabatically until its volume is doubled. The resultant fall in temperature of the gas is nearly (The ratio of the specific heats of the gas $\gamma = 1.5$) (in $K$)

If $\gamma$ denotes the ratio of two specific heats of a gas, the ratio of the slopes of adiabatic and isothermal $PV$ curves at their point of intersection is:

$A$ rigid diatomic gas $(\gamma = 7/5)$ is compressed adiabatically to volume $(V_i/32)$, where $V_i$ is the initial volume. The initial temperature of the gas is $T_i \text{ K}$ and the final temperature is $x T_i \text{ K}$. The value of $x$ is

$A$ monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ at a pressure of $4 \text{ atm}$ is compressed adiabatically so that its temperature rises from $27^{\circ} \text{C}$ to $327^{\circ} \text{C}$. The pressure of the gas in its final state is