Question 707: Thermodynamics & KTG

Question

A diatomic gas $\left(\gamma=\frac{7}{5}\right)$ is compressed adiabatically to volume $\frac{\mathrm{V}_0}{32}$, where $\mathrm{V}_0$ is its initial volume. The initial temperature of the gas is $\mathrm{T}_{\mathrm{i}}$ in kelvin and the final temperature is $\mathrm{xT}_{\mathrm{i}}$ in kelvin. The value of $x$ is

Accepted Answer

For an adiabatic process:
$$ TV^{\gamma-1} = 	ext{constant} $$
Let the initial state be $(T_i, V_0)$ and the final state be $(T_f, V_f)$.
Given:
$\gamma = \frac{7}{5}$
$V_f = \frac{V_0}{32}$
$T_f = xT_i$
We can write:
$$ T_i V_0^{\gamma-1} = T_f V_f^{\gamma-1} $$
Substitute the values:
$$ T_i V_0^{\gamma-1} = xT_i \left(\frac{V_0}{32}ight)^{\gamma-1} $$
Divide both sides by $T_i$:
$$ V_0^{\gamma-1} = x \left(\frac{V_0}{32}ight)^{\gamma-1} $$
Rewrite the right side:
$$ = x \cdot V_0^{\gamma-1} \cdot 32^{-(\gamma-1)} $$
Now, cancel $V_0^{\gamma-1}$ from both sides:
$$ 1 = x \cdot 32^{-(\gamma-1)} $$
Bring $32^{\gamma-1}$ to the left:
$$ x = 32^{\gamma-1} $$
Now, calculate $\gamma-1$:
$$ \gamma-1 = \frac{7}{5} - 1 = \frac{2}{5} $$
Now:
$$ x = 32^{2/5} $$
Recall, $32 = 2^5$:
$$ 32^{2/5} = (2^5)^{2/5} = 2^{2} = 4 $$
So,
$$ x = 4 $$
Correct Option: B (4)

Answer

5

Answer

4

Answer

3

Answer

2

Thermodynamics & KTG

Choose the correct answer: