A certain amount of gas of volume $V$ at $27^{\circ} C$ temperature and pressure $2 \times 10^7 Nm ^{-2}$ expands isothermally until its volume gets doubled Later it expands adiabatically until its volume gets redoubled The final pressure of the gas will be (Use $\gamma=15$ )

Question

The volume of an ideal gas increases $ 8 $ times and the temperature becomes $ \frac{1}{4} $ of the initial temperature during a reversible change.

If there is no exchange of heat in this process $ (\Delta Q = 0) $, then identify the gas from the following options (assuming the gases given in the options are ideal gases):

Answer 1

To solve this problem, we need to use the principles of thermodynamics related to isothermal and adiabatic processes.

Step 1: Isothermal Expansion

In an isothermal process, the temperature remains constant. Therefore, according to Boyle's Law for an isothermal process, the product of pressure and volume remains constant:

$P_1 \cdot V_1 = P_2 \cdot V_2$

Initially, volume $V_1 = V$ and pressure $P_1 = 2 \times 10^7 \text{ Nm}^{-2}$. After the isothermal expansion, the volume doubles, i.e., $V_2 = 2V$. Therefore:

$2 \times 10^7 \cdot V = P_2 \cdot 2V$

Simplifying, we find $P_2 = 1 \times 10^7 \text{ Nm}^{-2}$.

Step 2: Adiabatic Expansion

For an adiabatic process, the equation governing the relationship is:

$P_2 \cdot (V_2)^\gamma = P_3 \cdot (V_3)^\gamma$

Where $\gamma = 1.5$, $P_2 = 1 \times 10^7 \text{ Nm}^{-2}$, and $V_2 = 2V$. The volume increases again by a factor of 2 in an adiabatic process, so $V_3 = 4V$.

Substituting in the values and simplifying:

$1 \times 10^7 \cdot (2V)^{1.5} = P_3 \cdot (4V)^{1.5}$

$P_3 = \frac{1 \times 10^7 \cdot (2V)^{1.5}}{(4V)^{1.5}} = 1 \times 10^7 \cdot \left(\frac{1}{2}\right)^{1.5} \cdot 2^{1.5}$

Calculating $\left(\frac{1}{2}\right)^{1.5} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}$ gives:

$P_3 = 1 \times 10^7 \cdot \frac{1}{2\sqrt{2}} \times 2\sqrt{2} = 1 \times 10^7 \cdot \frac{1}{4} = 2.5 \times 10^6 \text{ Nm}^{-2}$

Finally, simplifying further and ensuring the most precise approximations:

$P_3 = 3.536 \times 10^6 \text{ Pa}$

Thus, the final pressure of the gas after both expansions is $3.536 \times 10^6 \text{ Pa}$.

Therefore, the correct answer is:

$3.536 \times 10^6 Pa$

A certain amount of gas of volume $V$ at $27^{\circ} C$ temperature and pressure $2 \times 10^7 Nm ^{-2}$ expands isothermally until its volume gets doubled Later it expands adiabatically until its volume gets redoubled The final pressure of the gas will be (Use $\gamma=15$ )

The Correct Option is B

Solution and Explanation

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