# 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?

Mar 18, 2017

PV = nRT

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

T = temperature in Kelvin

Therefore

$T = \frac{P V}{n R}$

Plug in the variables but 175kPa to atm

1kilopascal = 0.00986923atm

Thus 175kilopascal

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

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

= 194.690888736K