Answer:
1474.0 torr.
Explanation:
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant.
T is the temperature of the gas in K.
P₁V₁T₂ = P₂V₂T₁
P₁ = 895.0 torr, V₁ = 8.8 L, T₁ = 25°C + 273 = 298 K.
P₂ = ??? torr, V₂ = 5.9 L, T₂ = 56°C + 273 = 329 K.
∴ P₂ = P₁V₁T₂/V₂T₁ = (895.0 torr)(8.8 L)(329 K)/(5.9 L)(298 K) = 1474.0 torr.
Answer:
The pressure of the chlorine gas with volume equal 5.90 L at 56°C is 1,473.7 Torr.
Explanation:
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure chlorine gas = 895 Torr =[tex]\frac{895}{760}atm= 1.18 atm[/tex]
[tex]1 atm = 760 Torr[/tex]
[tex]P_2[/tex] = final pressure chlorine gas = ?
[tex]V_1[/tex] = initial volume chlorine gas =8.80 L
[tex]V_2[/tex] = final volume chlorine gas = 5.90 L
[tex]T_1[/tex] = initial temperature chlorine gas = [tex]25^oC=273.15+25=298.15 K[/tex]
[tex]T_2[/tex] = final temperature chlorine gas = [tex]56^oC=273.15+56=329.15 K[/tex]
Now put all the given values in the above equation, we get:
[tex]P_2=\frac{P_1V_1\times T_2}{T_1\times V_2}[/tex]
[tex]=\frac{1.18 atm\times 8.80 L\times 329.15 K}{298.15 K\times 5.90 L}[/tex]
[tex]P_2=1.94 atm=1.94\times 760 Torr=1,473.7 Torr[/tex]
The pressure of the chlorine gas with volume equal 5.90 L at 56°C is 1,473.7 Torr.
