Question

Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. suppose that the volume of a particular sample of cl2 gas is 8.80 l at 895 torr and 25 ∘c. at what pressure will the volume equal 5.90 l if the temperature is 56 ∘c?

Answer 1

The pressure of the chlorine gas with volume equal 5.90 L at 56°C is 1,473.7 Torr.

Step-by-step explanation:

The combined gas equation is,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure chlorine gas = 895 Torr =
`(895)/(760)atm= 1.18 atm`

`1 atm = 760 Torr`

`P_2` = final pressure chlorine gas = ?

`V_1` = initial volume chlorine gas =8.80 L

`V_2` = final volume chlorine gas = 5.90 L

`T_1` = initial temperature chlorine gas =
`25^oC=273.15+25=298.15 K`

`T_2` = final temperature chlorine gas =
`56^oC=273.15+56=329.15 K`

Now put all the given values in the above equation, we get:

`P_2=(P_1V_1* T_2)/(T_1* V_2)`

`=(1.18 atm* 8.80 L* 329.15 K)/(298.15 K* 5.90 L)`

`P_2=1.94 atm=1.94* 760 Torr=1,473.7 Torr`

The pressure of the chlorine gas with volume equal 5.90 L at 56°C is 1,473.7 Torr.

Answer 2

1474.0 torr.

Step-by-step explanation:

• To calculate the no. of moles of a gas, we can use the general law of ideal gas: PV = nRT.

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.

• If n is constant, and have two different values of (P, V and T):

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.