Calculate the volume of nitrogen gas produce by heating g of ammonia at 21c and 823 torr pressure

final volume = 10.5 Liters N₂(g) at 21°C and 823Torr*

Explanation:

*Note=>No specified mass value of N₂(g) is defined in the problem. Therefore for a starting point, the gas sample is assumed to be 1.00 mole N₂(g) at STP conditions  22.4L

Determine volume of  N₂(g) at 21°C(=294K) and 823 Torr (= 2.286 Atm).

Start with Volume of N₂(g) at 0°C and 1 Atm pressure => 22.4L and adjust to final volume of  N₂(g) based upon 21°C(=294K) and 823 Torr (= 2.286 Atm).

V(final) = 22.4L(294K/273K)(360 Torr/823 Torr) = 22.4L(294/273)(360/823) = 10.55 Liters final volume.

Note: The volume of 1 mole (assumed) of any gas at STP (0°C/1 Atm) is 22.4 Liters. To convert to non-STP conditions, convert temperature and pressure factors (changes) that reflect what happens when the gas is expanded or decreased; but, these adjustments are taken independently for each variable of interest.  The following notes explain.

For the increase in temperature from 0°C(=273K) to 21°C(=294K) one must apply a temperature ratio that will increase volume. That is, the change in volume due to the temperature change is 294K/273K. If a 273K/294K ratio were used the volume would have decreased. Not so for heating a sample of gas.

For the increase in pressure one should expect a decrease in volume. Therefore apply a pressure ration that will effectively decrease the volume of the gas. That is, to decrease a 22.4L sample at STP multiply the standard volume by a ratio of pressures that will decrease 22.4L to a smaller volume. That is, V(final by pressure effects) multiply by 360Torr/823Torr to decrease the STP VOLUME (22.4L) to the new non-standard volume. If 823Torr/360Torr were used, the final volume would not be smaller, but larger. Such is the physical effect of an increasing pressure change.

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