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# Answer to Question #86408 in General Chemistry for taha

Question #86408
A 9.87-L sample of gas has a pressure of 0.646 atm and a temperature of 87 &deg;C. The sample is allowed to expand to a volume of 11.0 L and is cooled to 31 &deg;C. Calculate the new pressure of the gas, assuming that no gas escaped during the experiment. _____ atm.
1
2019-03-18T08:13:46-0400

Let's denote the parameters of the gas in the initial condition by index 1, and in the final condition by index 2:

@$V_1 = 9.87~\text{L},~t_1 = 87~\degree\text{C},~P_1 = 0.646~\text{atm}; \\ V_2 = 11~\text{L},~t_2 = 31~\degree\text{C},~P_2 - ?@$

We are going to make use Combined gas law for the case when comparing the same substance under two different sets of conditions:

@$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}. @$

For the formula to be correct, the Celsius temperatures should be converted to absolute temperatures (in Kelvin):

@$T = (\frac{t}{\degree\text{C}} + 273.15)~\text{K}; \\ T_1 = (87 + 273.15)~\text{K} = 360.15~\text{K}; \\ T_2 = (31 + 273.15)~\text{K} = 304.15~\text{K}.@$

Solving the Combined gas law for the unknown pressure, and entering the numerical values,

@$P_2 = P_1\frac{V_1}{T_2}\frac{T_2}{V_1} = 0.646~\text{atm}~*~\frac{9.87~\cancel{\text{L}}}{360.15~\cancel{\text{K}}}~*~\frac{304.15~\cancel{\text{K}}}{11~\cancel{\text{L}}} \approx 0.489~\text{atm}.@$

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