Please use significant figures and show work.
A gas has a pressure of 699.0 mmHg at 40.0°C. What is the temperature at standard pressure?
Gay Lussac's Law. As volume isn't mentioned, I'll assume it's constant.
P1 x T2 = P2 x T1
699mmHg x T2 = 760mmHg x 313K (40°C)
T2 = 760 x 313 ÷ 699
Final Pressure = 237,880 ÷ 699 = 340.3K = 67.3°C.
Reply:Sorry about that but a slight error in my answer. I put 'Final Pressure' instead of Temperature. Report It
Reply:P1= 699.0 mm Hg= 93 kPa
T1= 40.0+273=313 K
P2=101.3 (standart pressure)
T2= X
93 divided by 313= .30X
101.3 divided by .30
temperature= 337 kelvin= 64 degrees celcius
i think this is right...
Reply:699.0 mm Hg / 313 K = 760 mm Hg / x
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