Question #df0fe

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Question #df0fe

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Answer 1 · 2016-07-06T20:29:02

It is $0.555$ $0.555$ atm.

Explanation:

The equation of the perfect gas is

$P V = n R T$ $PV=nRT$

$P$ $P$ is the pressure that we want to calculate, $V$ $V$ is the volume in $m^{3}$ $"m"^3$ , $n$ $n$ are the mols, $R$ $R$ is the constant of perfect gas $8.314472 \frac{m^{3} Pa}{mol K}$ $8.314472 ("m"^3"Pa")/("mol" "K")$ , $T$ $T$ is the temperature in $K$ $"K"$ .
First of all we need to transform our quantity in the proper units:

$V = 11.2 L = 11.2 \cdot 10^{- 3} m^{3}$ $V=11.2 "L"=11.2*10^-3 "m"^3$

$T = 30^{\circ} C = 30 + 273.15 K = 303.15 K$ $T=30^\circ"C"=30+273.15 "K"=303.15"K"$ .

The pressure is

$P=(nRT)/V=(0.25 cancel"mol"*8.314472 (cancel("m"^3)"Pa")/(cancel"mol" cancel"K")*303.15cancel"K")/(11.2*10^-3 cancel("m"^3))$

$=(0.25* 8.314472*303.15)/(11.2)*10^3"Pa"$

$=56261.88" Pa"$ , or, if we want in a more familiar unit, we can convert it in atm using the conversion $1"atm"=101325"Pa"$ then

$56261.88" Pa"= 56261.88/101325" atm"$

$=0.555" atm"$ .

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Question #df0fe

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