What is the temperature in K of 0.335 mole of neon in a 2.00 L vessel at 4.32 bar? R=0.08314 L bar/mol k

T = 308.57 K

Explanation:

Given data:

Number of moles of K = 0.335 mol

Volume of vessel = 2.00 L

Pressure in vessel = 4.32 bar

R = 0.08314 L.bar /mol.K

Temperature of K = ?

Solution:

Formula:

PV = nRT

T = PV/nR

By putting values,

T = 4.32 bar ×2.00 L /0.335 mol× 0.08314 L.bar /mol.K

T = 8.64 bar.L / 0.028 L.bar /K

T = 308.57 K

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