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A sample of nitrogen gas has a pressure of 7.58 kPa at 639 K. What will be the pressure at 311 K if the volume does not change?

Respuesta :

Over here you have to use the formula: P1/T1= P2/T2
P1= 7.58
T1= 639

P2= ?
T2= 311

7.58/639 = ?/311
= 3.689 kPa

Answer: 3.69 kPa

Explanation:

Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

[tex]P\propto T[/tex]     (At constant volume and number of moles)

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 7.58 kPa

[tex]P_2[/tex] = final pressure of gas = ?

[tex]T_1[/tex] = initial temperature of gas = 639 K

[tex]T_2[/tex] = final temperature of gas = 311 K

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]\frac{7.58 kPa}{639K}=\frac{P_2}{311K}[/tex]

[tex]P_2=3.69kPa[/tex]

Therefore, the final pressure of gas will be 3.69 kPa.

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