ideal gas equation dilemma please help

Oct 2016
2
0
australia
Hi everyone, I'd appreciate some help with this problem (it is driving me crazy)

Humpback whales have large lungs (volume of 3600 L). If a whale takes a full breath of air at the surface (pressure= 1.000 atm or 101.3 kPa) and the internal body temperature of the whale is 36.40°C, how many grams of oxygen gas are in the whales lungs? ( R= 8.314 kPa L/ K mol ; R= 0.0821 atm L/K mol).

what I did:
1- pV= nRt so I rearrange the formula to n=pv/Rt, chaged temperature in kelvin (from 36,4°C to 309.4K)

2- n= (1 atm×3600L )/ (0.0821atm/k mol × 309.4K) , so n = 141.74 mol

3- moles to grams: 141.74 mol × 32 g/mol of oxygen = 4535 g

So my final answer is 4535 and it appears to be wrong, i have done it a million times and I don't find the mistake I must have made. Please help me!:confused:
 

topsquark

Forum Staff
Jul 2013
56
1
Any place where Alyson Hannigan can find me
Hi everyone, I'd appreciate some help with this problem (it is driving me crazy)

Humpback whales have large lungs (volume of 3600 L). If a whale takes a full breath of air at the surface (pressure= 1.000 atm or 101.3 kPa) and the internal body temperature of the whale is 36.40°C, how many grams of oxygen gas are in the whales lungs? ( R= 8.314 kPa L/ K mol ; R= 0.0821 atm L/K mol).

what I did:
1- pV= nRt so I rearrange the formula to n=pv/Rt, chaged temperature in kelvin (from 36,4°C to 309.4K)

2- n= (1 atm×3600L )/ (0.0821atm/k mol × 309.4K) , so n = 141.74 mol

3- moles to grams: 141.74 mol × 32 g/mol of oxygen = 4535 g

So my final answer is 4535 and it appears to be wrong, i have done it a million times and I don't find the mistake I must have made. Please help me!:confused:
The only thing I can suggest is that your gas constant is slightly off. I used R = 0.082057 (L atm)/(K mol). That got me a result of 4537 g. (There is a link to a wikipedia article under the "R = " portion here.)

-Dan