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At what temperature will 1.30 mole of an ideal gas in a 1.30 l container exert a pressure of 1.05 atm?

Sagot :

The temperature of the ideal gas is 12.76 K.

We need to know about the ideal gas theory to solve this problem. The ideal gas is assumed that there is no interaction between particles in a gas. It can be determined by the equation

P . V = n . R . T

where P is pressure, V is volume, n is the number of moles gas, R is the ideal gas constant (8.31 J/mol.K) and T is temperature.

From the question above, we know that

n = 1.3 mol

V = 1.30 l = 1.3 x 10¯³ m³

P = 1.05 atm = 1.05 x 1.01 x 10⁵ Pa = 106050 Pa

By substituting the following parameters, we can calculate the temperature

P . V = n . R . T

106050 . 1.3 x 10¯³ = 1.3 . 8.31 . T

T = 137.865 / 10.803

T = 12.76 K

Hence, the temperature of the gas is 12.76 K.

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