Question
I need help with these two problems:
At a given pressure and temperature, it takes 4.55 min for a 1.5 L sample of He to effuse through a membrane. How long does it take for 1.5 L of H2 to effuse under the same conditions?
A gas-filled weather balloon with a volume of 58.7 L is released at sea-level conditions of 755 torr and 23°C. The balloon can expand to a maximum volume of 788 L. When the balloon rises to an altitude at which the temperature is -5°C and the pressure is 0.051 atm, will it reach its maximum volume?
I tried working this one out by following this equation T2*P1*V1/(T1*P2)*(1/760)= (268K*755 torr*58.7L)/(296K*0.051atm)*(1/760)=1035.25 L, but the answer is somehow wrong when I put it into the computer.
At a given pressure and temperature, it takes 4.55 min for a 1.5 L sample of He to effuse through a membrane. How long does it take for 1.5 L of H2 to effuse under the same conditions?
A gas-filled weather balloon with a volume of 58.7 L is released at sea-level conditions of 755 torr and 23°C. The balloon can expand to a maximum volume of 788 L. When the balloon rises to an altitude at which the temperature is -5°C and the pressure is 0.051 atm, will it reach its maximum volume?
I tried working this one out by following this equation T2*P1*V1/(T1*P2)*(1/760)= (268K*755 torr*58.7L)/(296K*0.051atm)*(1/760)=1035.25 L, but the answer is somehow wrong when I put it into the computer.
Answers
bobpursley
On the first, the law of diffusion is used: Effusion is inversely related to mole mass..
rateHe/rateH2= sqrt ( molmassH2)/sqrt(molemassH2)
in your case, rates is volume/time
rateHe/rateH2= sqrt ( molmassH2)/sqrt(molemassH2)
in your case, rates is volume/time
The problem with the second problem may be that what you are to enter is a yes or no answer and not the volume you have calculated.
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