Question

For a vessel at 1832 k containing 10 moles of steam at 50 atm volume would be

Answer 1

PV = nRT
50V = 10×1832×0.0821
V = (10×1832×0.0821)÷50
V= 30.08 L

Answer 2

For a vessel at 1832 k containing 10 moles of steam at 50 atm volume would be 30 L

Explanation:

For an ideal gas, we have the equation as follows :

$PV =nRT$

where,

P = pressure  = 50 atm

V = volume  = ?

R = Gas constant  = 0.0821Latm/molK

T = temperature  = 1832 k

n = no. of moles = 10

Putting in the values we get:

$50\times V=10\times 0.0821\times 1832$

$V=30L$

Thus the volume would be 30 Liters

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