A250. mL sample of gas at 1.00 atm and 20 degrees Celsius has the temperature increased to 40 degrees Celsius and the volume increased to 500. mL. What is the new pressure?

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A250. mL sample of gas at 1.00 atm and 20 degrees Celsius has the temperature increased to 40 degrees Celsius and the volume increased to 500. mL. What is the new pressure?

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Answer 1 · 2016-12-10T13:55:01

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

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$ , given a constant quantity of gas.

And thus, $\frac{P_{1} \times V_{1} \times T_{2}}{V_{2} \times T_{1}} = P_{2}$ $(P_1xxV_1xxT_2)/(V_2xxT_1)=P_2$ , even given this format, we can see the problem is dimensionally sound. We do need to use $absolute temperature.$ $"absolute temperature."$

$P_{2} = \frac{1.00 \cdot a t m \times 250 \cdot m L \times 313 \cdot K}{500 \cdot m L \times 293 \cdot K} ≅ \frac{1}{2} \cdot a t m .$ $P_2=(1.00*atmxx250*mLxx313*K)/(500*mLxx293*K)~=1/2*atm.$

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A250. mL sample of gas at 1.00 atm and 20 degrees Celsius has the temperature increased to 40 degrees Celsius and the volume increased to 500. mL. What is the new pressure?

1 Answer

Explanation:

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