General Chem. Q...need help

[biostudent101]

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hi,

if one mole of gas originally at STP is placed in a cantainer where the pressure is doubled and the absolute temperature(k) is tripled, what is the new volume in liters?

a.2.2 liters
b. 5.6 L
c. 7.5 L
d. 11.2 L
e. 33.6 L

the answer is choice e.
the rule says that if the pressure increase the volume decrease; also when the Temp. increases volume increases.

can someone explain how they derived to the answer.
Thanks.

 

[llungu]

PV = nRT

 

[watt]

try breaking up the problem into individual parts: pressure increase and temperature increase.

if pressure is doubled, ur volume is cut in half. Since 1 mole of any gas at STP is 22.4L/mol, and you have 1 mole, 22.4L/mol x 1 mol = 22.4L
22.4/2 = 11.2L @ double the pressure.

if you triple the temperature, volume also triples:
11.2L x 3 = 33.6L

Try posting in the DAT section.. u might get more indepth explanations

 

[porterhouse]

P1V1/T1 = P2V2/T2

The question is asking for the new volume (V2). So after rearranging the equation to solve for V2 we get,

V2 = P1V1T2/T1P2

The pressure is doubled à 2P2 and,
the temperature is tripled à 3T2 therefore,

V2 = P1V13T2/T12P2

So, the new volume, V2, is 3/2 of the initial volume of one mole of gas at STP (22.4 L), so we get,

V2 = (3/2)(22.4 L) = 33.6 L (choice e)

Hope this helps.