QuestionApril 30, 2025

(4) A 1.67 mathrm(~g) Sample of mathrm(He) gas has a pressure of 0.234 mathrm(~atm) and a volume of 35.8 mathrm(~L) What is the temperature of the Sample in ( )^circ mathrm(C) [ 273^circ mathrm(C) -118^circ mathrm(C) 300^circ mathrm(C) -29.0^circ mathrm(C) 118^circ mathrm(C) ]

(4) A 1.67 mathrm(~g) Sample of mathrm(He) gas has a pressure of 0.234 mathrm(~atm) and a volume of 35.8 mathrm(~L) What is the temperature of the Sample in ( )^circ mathrm(C) [ 273^circ mathrm(C) -118^circ mathrm(C) 300^circ mathrm(C) -29.0^circ mathrm(C) 118^circ mathrm(C) ]
(4) A 1.67 mathrm(~g) Sample of mathrm(He) gas has a pressure of 0.234 mathrm(~atm) and a volume of 35.8 mathrm(~L) What is the temperature of the Sample in ( )^circ mathrm(C) [ 273^circ mathrm(C) -118^circ mathrm(C) 300^circ mathrm(C) -29.0^circ mathrm(C) 118^circ mathrm(C) ]

Solution
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Answer

-34.0^{\circ} \mathrm{C} Explanation 1. Use Ideal Gas Law Apply PV = nRT. Calculate moles n using n = \frac{\text{mass}}{\text{molar mass}}. Molar mass of He is 4.00 \, \mathrm{g/mol}. 2. Calculate Moles n = \frac{1.67 \, \mathrm{g}}{4.00 \, \mathrm{g/mol}} = 0.4175 \, \mathrm{mol} 3. Solve for Temperature in Kelvin Rearrange to T = \frac{PV}{nR}. Use R = 0.0821 \, \mathrm{L \cdot atm/(mol \cdot K)}. T = \frac{0.234 \, \mathrm{atm} \times 35.8 \, \mathrm{L}}{0.4175 \, \mathrm{mol} \times 0.0821 \, \mathrm{L \cdot atm/(mol \cdot K)}} T = 239.15 \, \mathrm{K} 4. Convert Kelvin to Celsius T_{\mathrm{C}} = T_{\mathrm{K}} - 273.15 T_{\mathrm{C}} = 239.15 - 273.15 = -34.0^{\circ} \mathrm{C}

Explanation

1. Use Ideal Gas Law<br /> Apply $PV = nRT$. Calculate moles $n$ using $n = \frac{\text{mass}}{\text{molar mass}}$. Molar mass of He is $4.00 \, \mathrm{g/mol}$.<br />2. Calculate Moles<br /> $n = \frac{1.67 \, \mathrm{g}}{4.00 \, \mathrm{g/mol}} = 0.4175 \, \mathrm{mol}$<br />3. Solve for Temperature in Kelvin<br /> Rearrange to $T = \frac{PV}{nR}$. Use $R = 0.0821 \, \mathrm{L \cdot atm/(mol \cdot K)}$.<br /> $T = \frac{0.234 \, \mathrm{atm} \times 35.8 \, \mathrm{L}}{0.4175 \, \mathrm{mol} \times 0.0821 \, \mathrm{L \cdot atm/(mol \cdot K)}}$<br /> $T = 239.15 \, \mathrm{K}$<br />4. Convert Kelvin to Celsius<br /> $T_{\mathrm{C}} = T_{\mathrm{K}} - 273.15$<br /> $T_{\mathrm{C}} = 239.15 - 273.15 = -34.0^{\circ} \mathrm{C}$
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