5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard, what is the new pressure?

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5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard, what is the new pressure?

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Answer 1 · 2016-03-20T03:51:46

Use the Ideal Gas Laws to calculate the change. $\frac{P_{1} \cdot V_{1}}{T_{1}} = \frac{P_{2} \cdot V_{2}}{T_{2}}$ $(P_1 * V_1)/T_1 = (P_2 * V_2)/T_2$

Explanation:

STP is 0°C (273.15K)and 760mm Hg. The temperatures must be in K for the ratio to be correct.

$\frac{P_{1} \cdot V_{1}}{T_{1}} = \frac{P_{2} \cdot V_{2}}{T_{2}}$ $(P_1 * V_1)/T_1 = (P_2 * V_2)/T_2$

$\frac{745 \cdot 5.00}{295.15} = \frac{760 \cdot V_{2}}{273.15}; V_{2} = \frac{273.15}{760} \cdot [\frac{745 \cdot 5.00}{295.15}]$ $(745 * 5.00)/295.15 = (760 * V_2)/273.15 ; V_2 = 273.15/760 * [(745 * 5.00)/295.15]$

$V_{2} = 4.53 L$ $V_2 = 4.53L$

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5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard, what is the new pressure?

1 Answer

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