Answer:
3 °C
Explanation:
The pressure is constant, so this looks like a case where we can use Charles’ Law:
[tex]\frac{ V_{1} }{T_{1}} = \frac{V_{2} }{T_{2}}[/tex]
Invert both sides of the equation.
[tex]\frac{T_{1} }{V_{1}} = \frac{T_{2} }{V_{2}}[/tex]
Multiply both sides by V₂
[tex]T_{2} = T_{1} \times \frac{V_{2} }{ V_{1} }[/tex]
V₁ = 13.3 L; T₁ = (55 + 273.15) K = 328.15 K
V₂ = 11.2 L; T₂ = ?
[tex]T_{2} = \text{328.15 K} \times \frac{\text{11.2 L} }{\text{13.3 L}} = \text{276 K} = 3 ^{\circ}\text{C}[/tex]
Taking into account the Charles's law, the temperature of an 11.2 L sample of carbon monoxide, CO, at 744 torr is 276 K or 3 C.
First of all, you have to know that Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas.
This law says that when the amount of gas and pressure are kept constant, the quotient between the volume and the temperature will always have the same value:
[tex]\frac{V}{T}=k[/tex]
Studying two different states, an initial state 1 and a final state 2, it is satisfied:
[tex]\frac{V1}{T1}=\frac{V2}{T2}[/tex]
In this case, you can apply Charles's law because it can be seen that the pressure is constant, with a value of 744 torr at the beginning and end of the analysis. Then, you know:
Replacing in Charles's law:
[tex]\frac{11.2 L}{T1}=\frac{13.3 L}{328 K}[/tex]
Solving:
[tex]11.2 L=\frac{13.3 L}{328 K}xT1[/tex]
[tex]\frac{11.2 L}{\frac{13.3 L}{328 K}}=T1[/tex]
T₁= 276 K= 3 C
Finally, the temperature of an 11.2 L sample of carbon monoxide, CO, at 744 torr is 276 K or 3 C.
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