{V_i}\right) $

where $ n $ is the number of moles, $ R $ is the ideal gas constant, $ T $ is the temperature, $ V_f $ is the final volume, and $ V_i $ is the initial volume.

Gas A Expansion:

The pressure of gas $ A $ decreases by 50%, thus:

$ p_f = 0.5 p_i $

Using the ideal gas law $ pV = nRT $, we have:

$ p_f V_f = p_i V_i $

$ 0.5 p_i V_f = p_i V_i \Rightarrow V_f = 2 V_i, \text{ thus } \frac{V_f}{V_i} = 2 $

Therefore, the work done by gas $ A $ is:

$ W_A = nRT_1 \ln(2) $

Gas B Expansion:

The pressure of gas $ B $ decreases by 75%, thus:

$ p_f = 0.25 p_i $

Using the ideal gas law again:

$ p_f V_f = p_i V_i $

$ 0.25 p_i V_f = p_i V_i \Rightarrow V_f = 4 V_i, \text{ thus } \frac{V_f}{V_i} = 4 $

The work done by gas $ B $ is:

$ W_B = nRT_2 \ln(4) $

Determining the Temperature Ratio

Given that the work done by both gases is the same, $ W_A = W_B $, we equate the expressions for work:

$ nRT_1 \ln(2) = nRT_2 \ln(4) $

Solving for the temperature ratio:

$ T_1 \ln(2) = T_2 \ln(4) $

$ T_1 \ln(2) = T_2 \cdot 2 \ln(2) $

$ \frac{T_1}{T_2} = \frac{2 \ln(2)}{\ln(2)} = 2 $

Thus, the ratio of temperatures $ T_1 : T_2 $ is:

$ 2:1 $

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