Answer
$$p_2 = 120 \space kPa$$
Work Step by Step
1. Convert the temperature values to Kelvin scale:
$$T(K) = T(C^o) + 273.15$$ $$T_1(K) = (23) + 273.15 = 296.15$$ $$T_2(K) = (11) + 273.15 = 284.15$$
2. Use the Combined gas law to calculate the final pressure:
$$\frac{p_1V_1}{n_1T_1} = \frac{p_2V_2}{n_2T_2}$$
Assuming that $n$ and $V$ are constant:
$$\frac{p_1}{T_1} = \frac{p_2}{T_2}$$
Solving for the final pressure $p_2$:
$$\frac{p_1}{T_1} \times T_2 = p_2$$ $$p_2 = \frac{125 \space kPa}{296.15 \space K} \times 284.15 \space K = 120 \space kPa$$
