NCERT Class XI Chemistry States of Matter Solutions

Question : 12

Total: 23

Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar. (R = 0.083 bar dm3K–1mol–1)

Solution:

n = 4 moles, P = 3.32 bar, V = 5 dm3, R = 0.083 bar dm3K–1mol–1
We know that PV = nRT
T =

3.32×5

4×0.083

⇒ T =

16.6

0.332

= 50K

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