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A sample of argon at 300. °C and 50.0 atm pressure is cooled in the same container to a temperature of 0. °C. What is the new pressure, if the volume and amount of gas do not change?

Respuesta :

20.6031685 atm for the temperature

Answer: 105 atm

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

where,

[tex]P_1[/tex] = initial pressure of gas = 50 atm

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

[tex]T_1[/tex] = initial temperature of gas  = [tex]30^0c=(300+273)K=573K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]0^0c=(0+273)K=273K[/tex]

[tex]{50\times 573}={P_2\times 273}[/tex]

[tex]P_2=105atm[/tex]

Therefore, the new pressure of the gas will be 105 atm.