The pressure of two moles of an ideal gas at a temperature of 27 °C and volume 10\(^-2\)m\(^3\) is?
[R = 8.313 J mol\(^{-1}\)K\(^{-1}\)]

Anela

1 year ago

PV = nRT
P= ?
V=0.01
n= 2
R= 8.313
T= 27+273= 300

P0.01 = 2×8.313×300
=P0.01= 4987.8
DBS BY 0.01
P=4987.8/0.01
= 498780( changing to standard form =4.987×10⁵≈4.99×10⁵(approx)

Go_zii4

4 months ago

it's not clear
I don't see the solution

webtoon

1 year ago

no it is correct

tbosseni

2 years ago

my school is wrong, ols correct ot. the ans o
is 4.99 ×10^5

Henry1472

5 months ago

it is correct oo

ogunleyeferanmi5

4 years ago

I still not understand this logic you are using
I got 496.78 but how did you get urs