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
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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 = ?
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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.
