QuestionJune 20, 2025

What is the total pressure in a 10.0 Lflask which contains 0.127 mol of H_(2)(g) and 0.288 mol of N_(2)(g) at 20.0^circ C 0.681 atm 0.306 atm 0.998 atm 0.693 atm

What is the total pressure in a 10.0 Lflask which contains 0.127 mol of H_(2)(g) and 0.288 mol of N_(2)(g) at 20.0^circ C 0.681 atm 0.306 atm 0.998 atm 0.693 atm
What is the total pressure in a 10.0 Lflask which contains 0.127 mol of H_(2)(g) and 0.288 mol of N_(2)(g) at 20.0^circ C 0.681 atm 0.306 atm 0.998 atm 0.693 atm

Solution
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Answer

0.998 atm Explanation 1. Use Ideal Gas Law for Each Gas Calculate the pressure of each gas using PV = nRT. For H_2: P = \frac{nRT}{V} = \frac{0.127 \times 0.0821 \times 293}{10.0} atm. 2. Calculate Pressure for H_2 P_{H_2} = \frac{0.127 \times 0.0821 \times 293}{10.0} = 0.305 atm. 3. Calculate Pressure for N_2 Similarly, for N_2: P = \frac{0.288 \times 0.0821 \times 293}{10.0} atm. 4. Calculate Pressure for N_2 P_{N_2} = \frac{0.288 \times 0.0821 \times 293}{10.0} = 0.693 atm. 5. Add Partial Pressures Total pressure P_{total} = P_{H_2} + P_{N_2} = 0.305 + 0.693 atm.

Explanation

1. Use Ideal Gas Law for Each Gas<br /> Calculate the pressure of each gas using $PV = nRT$. For $H_2$: $P = \frac{nRT}{V} = \frac{0.127 \times 0.0821 \times 293}{10.0}$ atm.<br /><br />2. Calculate Pressure for $H_2$<br /> $P_{H_2} = \frac{0.127 \times 0.0821 \times 293}{10.0} = 0.305$ atm.<br /><br />3. Calculate Pressure for $N_2$<br /> Similarly, for $N_2$: $P = \frac{0.288 \times 0.0821 \times 293}{10.0}$ atm.<br /><br />4. Calculate Pressure for $N_2$<br /> $P_{N_2} = \frac{0.288 \times 0.0821 \times 293}{10.0} = 0.693$ atm.<br /><br />5. Add Partial Pressures<br /> Total pressure $P_{total} = P_{H_2} + P_{N_2} = 0.305 + 0.693$ atm.
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