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alexira [117]
3 years ago
10

I have an unknown volume of gas at a pressure of 0.5 atm and a temperature of 325k if I raise the pressure to 1.2 atm, decrease

the temperature to 320k, and measures the final volume to be 48 liters what was the initial volume of the gas?
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
7 0
Use the Equation of Clapeyron:

\frac{P_1.V_1}{T_1}=\frac{P_2.V_2}{T_2}\\
\\
\frac{0.5V_1}{235}=\frac{1.2*48}{320}\\
\\
320*0.5V_1=235*1.2*48\\
\\
V_1=\frac{235*1.2*48}{320*0.5}=84.5 liters
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If we substitute  ΔH^0_{vap}  for  \frac{q}{mass} in the above equation, we have;

specific heat capacity (c) = \frac{deltaH^0_{vap}}{T_{final}-T_{initial}}

Making (T_{final}- T_{initial}) the subject of the formula; we have:

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         = 227.76°C +93.0°C

          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C                

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