Question

question 2 (2 points)
what is the temperature of 2.00 moles of oxygen in 5.00 l at 2.00 atm and 60.9 k when the pressure was increased to 4.00 atm and the volume was reduced to 2.00 l?
this is a 1 type of problem so we use the equation 2.
2.00 mole represents both ( n_1 ) and 3, 5.00 l is 4, 2.00 atm is 5, 60.9 k is 6,
4.00 atm is 7, and 2.00 l is 8
rearrange the equation for 9, input the variables, then solve mathematically.
a. missing variable b. change in conditions c. density d. molar volume at stp
e. partial pressure f. ( 1 mol = 22.4 l ) g. ( d = m/v ) h. ( pv = nrt )
i. ( p_2v_2/n_2t_2 = p_1v_1/n_1t_1 ) j. ( p_{total}=p_1 + p_2+...... ) k. ( p ) l. ( p_1 ) m. ( p_2 )
n. ( v ) o. ( v_1 ) p. ( v_2 ) q. ( n ) r. ( n_1 ) s. ( n_2 ) t. ( r ) u. ( t )
v. ( t_1 ) w. ( t_2 ) x. 1.25 y. 12.5 z. 125

Explanation:

1. Since the problem involves a change in pressure, volume, and temperature conditions for a gas, it is a change in conditions problem.
2. The ideal gas law in the form of combined gas law (for change in conditions when amount of gas \(n\) is constant) is \(P_1V_1/T_1=P_2V_2/T_2\) or \(P_2V_2/n_2T_2 = P_1V_1/n_1T_1\) (when \(n_1 = n_2\), it simplifies).
3. Given that the amount of gas (moles) does not change, \(n_1=n_2\).
4. \(V_1\) is the initial volume.
5. \(P_1\) is the initial pressure.
6. \(T_1\) is the initial temperature.
7. \(P_2\) is the final pressure.
8. \(V_2\) is the final volume.
9. We are solving for the final temperature \(T_2\) (a missing variable in the context of the problem).

Answer:

1. B. change in conditions
2. I. \(P_2V_2/n_2T_2 = P_1V_1/n_1T_1\)
3. S. \(n_2\)
4. O. \(V_1\)
5. L. \(P_1\)
6. V. \(T_1\)
7. M. \(P_2\)
8. P. \(V_2\)
9. A. missing variable