The new volume of the gas at 20 °C is 205 mL
The correct answer to the question is Option C. 205 mL
From the question given, the following data were obtained:
Initial volume (V₁) = 250 mL
Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
Initial pressure (P₁) = 10 atm
Final temperature (T₂) = 20 °C = 20 + 273 = 293 K
Final pressure (P₂) = 12 atm
Final volume (V₂) =?
The new volume of the gas can be obtained by using the combine gas equation as follow:
[tex] \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \\ \\ \frac{10 \times 250}{298} = \frac{12 \times V_2}{293} \\ \\ cross \: multiply \\ \\ 298 \times 12 \times V_2 = 10 \times 250 \times 293 \\ \\ divide \: both \: side \: by \: 298 \times 12 \\ \\ V_2 = \frac{10 \times 250 \times 293}{298 \times 12} \\ \\ V_2 = 205 \: mL \\ [/tex]
Therefore, the new volume of gas is 205 mL
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