At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at the height where temperature is 10°C and volume of the gas 640 mL ?
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Given :
V1 = Initial volume = 600 mL,
V2 = Final volume = 640 mL
P1 = Initial pressure = 760 mm Hg
T1 = Initial temperature = 25°C
= 25 + 273.15
K = 298.15 K
T2 = Final temperature = 10°C
= 10 + 273.15
K = 283.15 K
P2 = Final pressure
Formula :
\(\frac{P_1V_1}{T_1}\) = \(\frac{P_2V_2}{T_2}\)
Calculation :
According to combined gas law,
\(\frac{P_1V_1}{T_1}\) = \(\frac{P_2V_2}{T_2}\)
∴ P2 = \(\frac{P_1V_1T_2}{T_1V_2}\)
= \(\frac{760\times 600\times 283.15}{298.15\times 640}\)
= 676.654 mm Hg
∴ The final pressure of a gas is 676.654 mm Hg.
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