Question 855: Thermodynamics & KTG

Question

An ideal gas at pressure '$$p$$' is adiabatically compressed so that its density becomes twice that of the initial. If $$\gamma=\frac{c_p}{c_v}=\frac{7}{5}$$, then final pressure of the gas is

Accepted Answer

Here's how to solve this problem:
For an adiabatic process, we have the following relationship:
$$p_1V_1^\gamma = p_2V_2^\gamma$$
where:
*  $$p_1$$ and $$V_1$$ are the initial pressure and volume.
*  $$p_2$$ and $$V_2$$ are the final pressure and volume.
*  $$\gamma$$ is the adiabatic index (ratio of specific heats).
We're given that the density doubles. Density is mass per unit volume, so if the density doubles, the volume is halved. Therefore, $$V_2 = \frac{1}{2}V_1$$.
Substituting these values into the adiabatic equation:
$$p_1V_1^\gamma = p_2 \left(\frac{1}{2}V_1ight)^\gamma$$
Simplifying:
$$p_1 = p_2 \left(\frac{1}{2}ight)^\gamma$$
Solving for $$p_2$$:
$$p_2 = p_1 \left(\frac{1}{2}ight)^{-\gamma} = p_1 	imes 2^\gamma$$
Substituting the given value of $$\gamma = \frac{7}{5}$$:
$$p_2 = p_1 	imes 2^{7/5} \approx p_1 	imes 2.63$$
Therefore, the final pressure of the gas is approximately
2.63p
. The correct answer is
Option D
.

Answer

p

Answer

2p

Answer

$$\frac{7}{5}$$p

Answer

2.63p

Thermodynamics & KTG

An ideal gas at pressure '$p' is adiabatically compressed so that its density becomes twice that of the initial. If \gamma=\frac{c_p}{c_v}=\frac{7}{5}$, then final pressure of the gas is

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