A glass container was initially charged with 2.00 moles of a gas sample at 3.75 atm and 21.7 °C. Some of the gas was released as the temperature was increased to 28.1 °C, so the final pressure in the container was reduced to 0.998 atm. How many moles of the gas sample are present at the end?

Question

Makenna Collier

Chemistry

10 July, 23:27

A glass container was initially charged with 2.00 moles of a gas sample at 3.75 atm and 21.7 °C. Some of the gas was released as the temperature was increased to 28.1 °C, so the final pressure in the container was reduced to 0.998 atm. How many moles of the gas sample are present at the end?

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Answers (1)

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Jase · Answer 1

0.521 moles still present in the container.

Explanation:

It is possible to answer this question by using the general gas law, that is:

PV = nRT

Where P represents pressure of the gas, v its volume, n moles, R gas constant law and T absolute temperature (21.7°C + 273.15 = 294.85K)

Replacing with values of the initial conditions of the container, its volume is:

V = nRT / P

V = 2.00mol*0.082atmL/molK*294.85K / 3.75atm

V = 12.9L

When some gas is released, absolute temperature is 28.1°C + 273.15 = 301.25K, the pressure is 0.998atm and the volume of the container still constant. Again, using general gas law:

PV / RT = n

0.998atm*12.9L / 0.082atmL/molK*301.25K = n

0.521 moles = n

0.521 moles still present in the container.