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A 1.13 mole sample of a gas at 2.6 atm is in a 18.4 L container. What is the temperature in Kelvin inside the container?

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To solve this question, you need to use Clapeyron's equation:

P.V = n.R.T

P = pressure generated by the gas on the walls of the container;

V = volume occupied by the gas and can be expressed in liters or cubic meters;

n = number of mol (amount of substance in the gas);

R = general gas constant proposed by Clapeyron and depends on the pressure unit used (in atm, it is 0.082; in mmHg, it is 62.3; in KPa, it is 8.31);

T = temperature at which the gas is subjected (always used in the unit Kelvin).

So we have:

n = 1.13

P = 2.6 atm

V = 18.4 L

T = ?

R = 0.082 atm.L/mol.K

PV = nRT

2.6 x 18.4 = 1.13 x 0.082 x T

47.84 = 0.09266T

T = 516.3 K

Answer: The temperature is 516.3 K

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