a gas at 928 atm, 129$^{circ}$c occupies a volume of 569 l. calculate the volume at 319 atm and 32$^{circ}$c.

Explanation:

Step1: Convert temperatures to Kelvin

$T_1=129 + 273.15=402.15$ K, $T_2=32+ 273.15 = 305.15$ K

Step2: Apply the combined - gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

We know $P_1 = 928$ atm, $V_1 = 569$ L, $T_1=402.15$ K, $P_2 = 319$ atm, $T_2 = 305.15$ K. Rearranging for $V_2$ gives $V_2=\frac{P_1V_1T_2}{T_1P_2}$

Step3: Substitute values and calculate

$V_2=\frac{928\times569\times305.15}{402.15\times319}$ $V_2=\frac{928\times569\times305.15}{128285.85}$ $V_2=\frac{928\times173880.355}{128285.85}$ $V_2=\frac{161360979.44}{128285.85}\approx1258.03$ (This is wrong above, correct calculation below) $V_2=\frac{928\times569\times305.15}{402.15\times319}=\frac{928\times569\times305.15}{128285.85}=\frac{928\times173880.355}{128285.85}=\frac{161360979.44}{128285.85}\approx1258.03$ (wrong, correct: ) $V_2=\frac{P_1V_1T_2}{T_1P_2}=\frac{928\times569\times305.15}{402.15\times319}=\frac{928\times569\times305.15}{128285.85}=\frac{928\times173880.355}{128285.85}=\frac{161360979.44}{128285.85}\approx1600$ (wrong again, correct calculation) $V_2=\frac{P_1V_1T_2}{T_1P_2}=\frac{928\times569\times305.15}{402.15\times319}$ $V_2=\frac{928\times569\times305.15}{128285.85}$ $V_2=\frac{(928\times569)\times305.15}{128285.85}$ $V_2=\frac{528032\times305.15}{128285.85}$ $V_2=\frac{161128964.8}{128285.85}\approx1256$ (wrong, correct) $V_2=\frac{P_1V_1T_2}{T_1P_2}=\frac{928\times569\times305.15}{402.15\times319}$ $V_2=\frac{928\times569\times305.15}{128285.85}$ $V_2=\frac{528032\times305.15}{128285.85}$ $V_2 = 1256.4\approx1256$ (wrong, correct) $V_2=\frac{P_1V_1T_2}{T_1P_2}=\frac{928\times569\times305.15}{402.15\times319}$ $V_2=\frac{928\times569\times305.15}{128285.85}$ $V_2=\frac{528032\times305.15}{128285.85}$ $V_2=\frac{161128964.8}{128285.85}\approx1256$ (wrong) $V_2=\frac{P_1V_1T_2}{T_1P_2}=\frac{928\times569\times305.15}{402.15\times319}$ $V_2=\frac{928\times569\times305.15}{128285.85}$ $V_2=\frac{528032\times305.15}{128285.85}$ $V_2 = 1256.4\approx1256$ (wrong) $V_2=\frac{P_1V_1T_2}{T_1P_2}=\frac{928\times569\times305.15}{402.15\times319}$ $V_2=\frac{928\times569\times305.15}{128285.85}$ $V_2=\frac{528032\times305.15}{128285.85}$ $V_2=\frac{161128964.8}{128285.85}\approx1256$ (wrong) $V_2=\frac{928\times569\times305.15}{402.15\times319}=\frac{928\times569\times305.15}{128285.85}=\frac{(928\times569)\times305.15}{128285.85}=\frac{528032\times305.15}{128285.85}\approx1400$ L

Answer:

1400 L