If 0.40 mol of a gas in a 3.7 L container is held at a pressure of 175 kPa, what is the temperature of the gas?

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If 0.40 mol of a gas in a 3.7 L container is held at a pressure of 175 kPa, what is the temperature of the gas?

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Answer 1 · 2017-03-18T10:04:49

PV = nRT

Where
P = pressure
V = volume
n = moles
R is the gas constant with a value of
= $0.082057338 L atm K^{- 1} m o l^{- 1}$ $"0.082057338 L atm" K^-1 mol^-1$

T = temperature in Kelvin

Therefore

$T = \frac{P V}{n R}$ $T = (PV)/(nR)$

Plug in the variables but 175kPa to atm

1kilopascal = 0.00986923atm

Thus 175kilopascal

= $175 k P a \cdot 0.00986923 a t m = 1.72711525 a t m$ $175kPa * 0.00986923atm = 1.72711525atm$

$T = \frac{1.72711525 atm \cdot 3.7 L}{0.4 m o l \cdot 0.082057338 L atm K^{- 1} m o l^{- 1}}$ $T= (1.72711525"atm" * 3.7L)/(0.4mol*"0.082057338 L atm" K^-1 mol^-1)$

= 194.690888736K

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If 0.40 mol of a gas in a 3.7 L container is held at a pressure of 175 kPa, what is the temperature of the gas?

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