An ideal gas at atmospheric pressure is adiabatically compressed so that its density becomes $32$ times of its initial value. If the final pressure of gas is $128$ atmospheres, the value of ?$\gamma$?of the gas is :

Question

An insulated cylinder of volume $60\,\text{cm}^3$ is filled with a gas at $27^\circ\text{C}$ and $2$ atmospheric pressure. The gas is then compressed making the final volume $20\,\text{cm}^3$ while allowing the temperature to rise to $77^\circ\text{C}$. The final pressure is ____________ atmospheric pressure.

Answer 1

To find the value of $\gamma$ for the gas, we need to apply the concept of adiabatic processes in thermodynamics. The relation for an adiabatic process can be expressed as:

$ P V^{\gamma} = \text{constant}$.

We also know that $ P = \rho R T $ for ideal gases, where $\rho$ is the density. Thus, from the given problem specifics:

The initial pressure is $ P_1 = 1 $ atm (atmospheric pressure).

The gas is compressed such that its density increases $ \rho_2 = 32 \rho_1 $.

The final pressure is $ P_2 = 128 $ atm.

The relation for adiabatic compression can also be expressed using densities and pressures:

$ \frac{P_2}{P_1} = \left(\frac{\rho_2}{\rho_1}\right)^{\gamma} $.

Substitute the given values:

$ \frac{128}{1} = (32)^{\gamma} $.

Therefore,

$ 128 = 32^{\gamma} $.

Rewriting 128 as a power of 2, we have $ 128 = 2^7 $ and $ 32 = 2^5 $.

Hence, the equation becomes $ 2^7 = (2^5)^{\gamma} $ or $ 2^7 = 2^{5\gamma} $.

By comparing exponents, we get:

$ 7 = 5\gamma $.

Solving for $\gamma$, we find:

$ \gamma = \frac{7}{5} = 1.4 $.

Thus, the correct answer is $1.4$, which matches the provided correct option.

Answer 2