Explanation :
As we know that the Gibbs free energy is not only function of temperature and pressure but also amount of each substance in the system.

where,
is the amount of component 1 and 2 in the system.
Partial molar Gibbs free energy : The partial derivative of Gibbs free energy with respect to amount of component (i) of a mixture when other variable
are kept constant are known as partial molar Gibbs free energy of
component.
For a substance in a mixture, the chemical potential
is defined as the partial molar Gibbs free energy.
The expression will be:

where,
T = temperature
P = pressure
is the amount of component 'i' and 'j' in the system.
Answer:
<u>1</u><u>.</u><u> </u><u>0</u><u>0</u><u>0</u><u>.</u><u>2</u>
Explanation:
<u> </u><u> </u><u>a</u><u>l</u><u>u</u><u>m</u><u>i</u><u>n</u><u>u</u><u>m</u><u> </u><u>o</u><u>x</u><u>i</u><u>d</u><u>e</u><u> </u><u>m</u><u>a</u><u>d</u><u>e</u><u> </u><u>o</u><u>f</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>r</u><u> </u><u>a</u><u>t</u><u>o</u><u>m</u><u>s</u><u> </u><u>l</u><u>e</u><u>s</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u>n</u><u> </u><u>o</u><u>t</u><u>h</u><u>e</u><u>r</u><u> </u><u>e</u><u>l</u><u>e</u><u>m</u><u>e</u><u>n</u><u>t</u><u>s</u><u> </u><u>n</u><u>e</u><u>e</u><u>d</u><u> </u><u>t</u><u>o</u><u> </u><u>d</u><u>e</u><u>c</u><u>r</u><u>e</u><u>a</u><u>s</u><u>e</u>
Answer:
The new temperature of the nitrogen gas is 516.8 K or 243.8 C.
Explanation:
Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.
Gay-Lussac's law can be expressed mathematically as follows:
Where P = pressure, T = temperature, K = Constant
You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

In this case:
- P1= 2 atm
- T1= 50 C= 323 K (being 0 C= 273 K)
- P2= 3.2 atm
- T2= ?
Replacing:

Solving:


T2= 516.8 K= 243.8 C
<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>