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kodGreya [7K]
3 years ago
12

Suppose that the volume of a particular sample of Cl2 is 718 mL at 675 mmHg and 48C at what temperature in C will the volume be

2.0 L if the pressure is 159 kPa
Chemistry
1 answer:
Kryger [21]3 years ago
4 0

Answer:

1900 °C

Step-by-step explanation:

This looks like a case where we can use the <em>Combined Gas Law</em> to calculate the temperature.

p₁V₁/T₁ = p₂V₂/T₂               Multiply both sides by T₂

p₁V₁T₂/T₁ = p₂V₂                 Multiply each side by T₁

p₁V₁T₂ = p₂V₂T₁                   Divide each side by p₁V₁

T₂ = T₁ × p₂/p₁ × V₂/V₁  

=====

Data:

We must convert the pressures to a common unit. I have chosen atmospheres.

p₁ = 675 mmHg × 1atm/760 mmHg = 0.8882 atm

V₁ = 718 mL = 0.718 L

T₁ = 48 °C = 321.15 K

p₂ = 159 kPa × 1 atm/101.325 kPa = 1.569 atm

V₂ = 2.0 L

T₂ = ?

=====

Calculation:

T₂ = 321.15 × 1.569/0.8882 × 2.0/0.718  

T₂ = 321.15 × 1.766 × 2.786  

T₂ = 321.15 × 1.569/0.8882 × 7.786  

T₂ = 1580K  

T₂ = 1580 + 273.15

T₂ = 1900 °C

<em>Note</em>: The answer can have only <em>two</em> significant figures because that is all you gave for the second volume of the gas.

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An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder
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Answer:

\Delta H=-11897J

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

\Delta H=\Delta U+V\Delta P

Whereas the change in the internal energy is computed by:

\Delta U=nCv\Delta T

So we compute the initial and final temperatures for one mole of the ideal gas:

T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K  }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K  }{n}

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J

Then, the volume-pressure product in Joules:

V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J

Finally, the change in the enthalpy for the process:

\Delta H=-7140J-4757J\\\\\Delta H=-11897J

Best regards.

7 0
3 years ago
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