If a sample of N gas was put into a flexible container at an pressure of 25.5kPa, temperature of 35.9 C and a volume of 500. mL, what would the final temperature of the gas be if the pressure was changed to .625 atm and the volume was change to 625mL?

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If a sample of N gas was put into a flexible container at an pressure of 25.5kPa, temperature of 35.9 C and a volume of 500. mL, what would the final temperature of the gas be if the pressure was changed to .625 atm and the volume was change to 625mL?

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Answer 1 · 2017-03-05T12:23:20

$705^{\circ} C$ $705 ^@"C"$

Explanation:

First look at the things that need to be converted

${35.9}^{\circ} C = (35.9 + 273.15) K = 309.0 K$ $35.9 ^@"C" = "(35.9 + 273.15) K" = "309.0 K"$

$25.5 kPa = 0.2467 atm$ $"25.5 kPa = 0.2467 atm"$

$500 mL = 0.500 L$ $"500 mL = 0.500 L"$

$625 mL = 0.625 L$ $"625 mL = 0.625 L"$

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1 = (P_2V_2)/T_2$

Plug in the variables

$\frac{0.2467 atm \times 0.500 L}{309.0 K} = \frac{0.625 atm \times 0.625 L}{T_{2}}$ $("0.2467 atm" xx "0.500 L")/("309.0 K") = ("0.625 atm" xx "0.625 L")/T_2$

$T_{2} = 978.5 K$ $T_2 = 978.5K$

or $T_{2} = {(978.5 - 273.15)}^{\circ} C = 705^{\circ} C$ $T_2 = (978.5 - 273.15) ^@"C" = 705 ^@"C"$

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If a sample of N gas was put into a flexible container at an pressure of 25.5kPa, temperature of 35.9 C and a volume of 500. mL, what would the final temperature of the gas be if the pressure was changed to .625 atm and the volume was change to 625mL?

1 Answer

Explanation:

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