X moles of nitrogen gas at STP conditions occupy a volune of 10 litres then the volume of 2x moles of CH4 at 273 degree centigrade 1.5 atm is?

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X moles of nitrogen gas at STP conditions occupy a volune of 10 litres then the volume of 2x moles of CH4 at 273 degree centigrade 1.5 atm is?

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X moles of nitrogen gas at STP conditions occupy a volune of 10 litres...

P = pressure

V = volume

n = number of moles

R = ideal gas constant

T = temperature (in Kelvin)

S.T.P = 1atm at 0 degrees Celsius

No I do have to ask you, are you certain it's 273 Celsius or is it 273 Kelvin? 273 Kelvin is 0 degrees Celsius (roughly), and it would make sense that they gave you Kelvin temperature since that's what the Ideal Gas Law deals with.

1 * 10 = x * R * 273

10 = x * R * 273

R = 10 / (x * 273)

If you meant 273 K

1.5 * V = 2x * R * 273

(3/2) * V = 2x * (10 / (273x)) * 273

(3/2) * V = 20

V = 20 * (2/3)

V = 40/3

V = 13.333333 Liters

If you meant 273 C

(3/2) * V = 2x * (273 + 273) * 10 / (273x)

(3/2) * V = 20 * 2

V = 40 * (2/3)

V = 80/3

V = 26.666666 Liters

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