Asked by Thao
A 500 mL metal cylinder holding 0.100 moles of helium gas is known to rupture at a pressure of 10 atmospheres. At what temperature, in °C, will the container fail?
Here is what I came up with:
T= (1.00mol x 0.0821 LxATM/KxMOL) / 10atm x 0.5L
=0.01642K
Convert to C: 0.01642K - 273 = -272.98 C
That can't possibly be right. Please help.
Here is what I came up with:
T= (1.00mol x 0.0821 LxATM/KxMOL) / 10atm x 0.5L
=0.01642K
Convert to C: 0.01642K - 273 = -272.98 C
That can't possibly be right. Please help.
Answers
Answered by
Harpoon
No that is correct the answer will turn out to be a negative number.
Answered by
Thao
I was giving the answer key with the correct answer at 336 C.
Now i'm really confused... Can someone explain.
Now i'm really confused... Can someone explain.
Answered by
DrBob222
Note that you substituted 1.00 mole for n and not 0.100 mol; also, you inverted the formula when you solved for T.
PV = nRT and rearrange to
T = PV/nR = (10)(0.500)/(0.100)(0.08206) = 609.3 K
609.3-273.15 = 336.16 which rounds to 336 C.
PV = nRT and rearrange to
T = PV/nR = (10)(0.500)/(0.100)(0.08206) = 609.3 K
609.3-273.15 = 336.16 which rounds to 336 C.
Answered by
Thao
I recalculated the problem and got it correct. Thank you.
Answered by
it is 336
stupid responder
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