Answer :
Answer:
a
The pressure will increase
b
[tex]T_2 = 576^oC[/tex]
Explanation:
From the ideal gas law we have that
[tex]PV = nRT[/tex]
We see that the temperature varies directly with the pressure so if there is an increase in temperature that pressure will increase
The initial temperature is [tex]T_i = 10^oC = 10 + 273 = 283 \ K [/tex]
The objective of this solution is to obtain the temperature of the gas where the pressure is tripled
Now from the above equation given that nR and V are constant we have that
[tex]\frac{P}{T} = constant[/tex]
=> [tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]
Let assume the initial pressure is [tex]P_1 = 1 Pa[/tex]
So tripling it will result to the pressure being [tex]P_2 = 3 Pa[/tex]
So
[tex]\frac{1}{283} =\frac{3}{T_2}[/tex]
=> [tex]T_2 = 3 * 283[/tex]
=> [tex]T_2 = 3 * 283[/tex]
=> [tex]T_2 = 849 \ K [/tex]
Converting back to [tex]^oC[/tex]
[tex]T_2 = 849 - 273[/tex]
=> [tex]T_2 = 576^oC[/tex]