QuestionAugust 11, 2025

4. The volume of a gas is 27.5 mL at 22^circ C and 0.974 atm. What will the volume be at 15.0^circ C and 0.993 atm? (Hint: Remember to convert temperature to Kelvin using the equation T_((K))=T_((^circ C))+273)

4. The volume of a gas is 27.5 mL at 22^circ C and 0.974 atm. What will the volume be at 15.0^circ C and 0.993 atm? (Hint: Remember to convert temperature to Kelvin using the equation T_((K))=T_((^circ C))+273)
4. The volume of a gas is 27.5 mL at 22^circ C and 0.974 atm. What will the volume be at 15.0^circ C and 0.993 atm? (Hint: Remember to convert temperature to Kelvin using the equation T_((K))=T_((^circ C))+273)

Solution
4.7(248 votes)

Answer

26.7 mL Explanation 1. Convert Temperatures to Kelvin T_1 = 22 + 273 = 295 \, K, T_2 = 15 + 273 = 288 \, K 2. Apply Combined Gas Law Use **\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}**. Substitute known values: \frac{0.974 \times 27.5}{295} = \frac{0.993 \times V_2}{288}. 3. Solve for V_2 Rearrange and solve: V_2 = \frac{0.974 \times 27.5 \times 288}{295 \times 0.993} 4. Calculate the Final Volume Compute: V_2 \approx 26.7 \, mL

Explanation

1. Convert Temperatures to Kelvin<br /> $T_1 = 22 + 273 = 295 \, K$, $T_2 = 15 + 273 = 288 \, K$<br /><br />2. Apply Combined Gas Law<br /> Use **$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$**. Substitute known values: $\frac{0.974 \times 27.5}{295} = \frac{0.993 \times V_2}{288}$.<br /><br />3. Solve for $V_2$<br /> Rearrange and solve: $V_2 = \frac{0.974 \times 27.5 \times 288}{295 \times 0.993}$<br /><br />4. Calculate the Final Volume<br /> Compute: $V_2 \approx 26.7 \, mL$
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