One mole of an ideal gas expands adiabatically from an initial state (T_A, V₀) to final state (T_f, 5V₀). Another mole of the same gas expands isothermally from a different initial state (T_B, V₀) to the same final state (T_f, 5V₀). The ratio of the specific heats at constant pressure and constant volume of this ideal gas is γ. What is the ratio T_A/T_B?

The correct option is 5^γ−1.

Let us break down the problem step-by-step by analyzing the two processes described for the ideal gas.

Step 1: Adiabatic expansion of the first mole of gas
The first mole of gas expands adiabatically from an initial state with temperature T_A and volume V₀ to a final state with temperature T_f and volume 5V₀.
For an adiabatic process involving an ideal gas, the relationship between temperature T and volume V is given by:

$T·V^{\mathrm{\nu <mo>-</mo><mn>1</mn>}}=\text{constant}$

Applying this relation between the initial state (T_A, V₀) and final state (T_f, 5V₀):

$T_A·{\left(V_0\right)}^{\mathrm{\gamma <mo>-</mo><mn>1</mn>}}=T_f·{\left(5V_0\right)}^{\mathrm{\gamma <mo>-</mo><mn>1</mn>}}$

Rearranging this equation to express T_A in terms of T_f:

$T_A=T_f·{\left(\frac{5V_0}{V_0}\right)}^{\mathrm{\gamma <mo>-</mo><mn>1</mn>}}=T_f·5^{\mathrm{\gamma <mo>-</mo><mn>1</mn>}}$

Step 2: Isothermal expansion of the second mole of gas
The second mole of the same gas expands isothermally from an initial state with temperature T_B and volume V₀ to the same final state with temperature T_f and volume 5V₀.
In an isothermal process, the temperature of the gas remains constant throughout the process. Therefore, the initial temperature must equal the final temperature:

$T_B=T_f$

Step 3: Calculating the ratio T_A / T_B
Now, we take the ratio of T_A to T_B using the expressions derived above:

$\frac{T_A}{T_B}=\frac{T_f·5^{\mathrm{\gamma <mo>-</mo><mn>1</mn>}}}{T_f}=5^{\mathrm{\gamma <mo>-</mo><mn>1</mn>}}$

Thus, the ratio T_A/T_B is 5^γ−1.