In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be n times the initial pressure. The value of n is :

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Heat and Thermodynamics question

2020 · 5 Sep · Shift 2 · Q43

Heat and Thermodynamics question

2020 · 5 Sep · Shift 2 · Q43

View written solutionFree

Correct answer: A

For an adiabatic process of an ideal gas,

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

Since density $\rho = \dfrac{m}{V}$ and mass is constant,

$V \propto \frac{1}{\rho}$

So we can rewrite the adiabatic relation in terms of density:

$P\left(\frac{1}{\rho}\right)^\gamma = \text{constant}$

which gives

$P \propto \rho^\gamma$

Hence,

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

For a diatomic gas,

$\gamma = \frac{C_p}{C_v} = \frac{7}{5}$

Given:

$\frac{\rho_2}{\rho_1} = 32 = 2^5$

Therefore,

$\frac{P_2}{P_1} = 32^{7/5} = (2^5)^{7/5} = 2^7 = 128$

So,

$n = 128$

Checking options:

A: $128$ ✅
B: $32$ ❌
C: $326$ ❌
D: $\dfrac{1}{32}$ ❌

Therefore, the correct answer is A.

More from Heat and Thermodynamics

Sign in

Heat and Thermodynamics question

2020 · 5 Sep · Shift 2 · Q43

Heat and Thermodynamics question

2020 · 5 Sep · Shift 2 · Q43

View written solutionFree

Correct answer: A

For an adiabatic process of an ideal gas,

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

Since density $\rho = \dfrac{m}{V}$ and mass is constant,

$V \propto \frac{1}{\rho}$

So we can rewrite the adiabatic relation in terms of density:

$P\left(\frac{1}{\rho}\right)^\gamma = \text{constant}$

which gives

$P \propto \rho^\gamma$

Hence,

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

For a diatomic gas,

$\gamma = \frac{C_p}{C_v} = \frac{7}{5}$

Given:

$\frac{\rho_2}{\rho_1} = 32 = 2^5$

Therefore,

$\frac{P_2}{P_1} = 32^{7/5} = (2^5)^{7/5} = 2^7 = 128$

So,

$n = 128$

Checking options:

A: $128$ ✅
B: $32$ ❌
C: $326$ ❌
D: $\dfrac{1}{32}$ ❌

Therefore, the correct answer is A.