Respuesta :
Answer : The number of moles of gas added to the container were, 3.078 mole
Explanation :
Using ideal gas equation,
[tex]PV=nRT[/tex]
At constant volume the formula will be,
[tex]\frac{P_1}{P_2}=\frac{n_1\times T_1}{n_2\times T_2}[/tex]
where,
[tex]n_1[/tex] = initial moles of gas = 1.94 moles
[tex]n_2[/tex] = final moles of gas = ?
[tex]P_1[/tex] = initial pressure of gas = 450 torr
[tex]P_2[/tex] = final pressure of gas = 750 torr
[tex]T_1[/tex] = initial temperature of gas = [tex]25^oC=273+25=298K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]40^oC=273+40=313K[/tex]
Now put all the given values in the above formula, we get:
[tex]\frac{450torr}{750torr}=\frac{1.94mole\times 298K}{n_2\times 313K}[/tex]
[tex]n_2=3.078mole[/tex]
Therefore, the number of moles of gas added to the container were, 3.078 mole
Answer:
3.078 moles
Explanation:
Given:
Number of moles of gas, n = 1.94 moles
Initial temperature, T₁ = 25° C = 298 K
Initial pressure, P₁ = 450 torr
Final temperature, T₂ = 40° C = 313 K
Final pressure, P₂ = 750 torr
now,
we know
PV = nRT
where,
P is the pressure
V is the volume
n is the number of moles
R is the universal gas constant
T is the temperature
now for the initial stage, the above relation comes as:
450 × V = 1.94 × R × 298 .............(1)
for the final stage
750 × V = n × R × 313 ............... (2)
on dividing the equation 2 by 1 we get
[tex]\frac{750}{450}=\frac{n\times313}{1.94\times298}[/tex]
or
n = 3.078 moles
