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

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,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure chlorine gas = 895 Torr =$\frac{895}{760}atm= 1.18 atm$

$1 atm = 760 Torr$

$P_2$ = 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:

$P_2=\frac{P_1V_1\times T_2}{T_1\times V_2}$

$=\frac{1.18 atm\times 8.80 L\times 329.15 K}{298.15 K\times 5.90 L}$

$P_2=1.94 atm=1.94\times 760 Torr=1,473.7 Torr$

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

Answer 2

Answer:

1474.0 torr.

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.