Question

The volume of a sample of Oxygen is 0.3L when the pressure is 1 atm and the temperature is 300K. At what temperature is the volume at 1L and the pressure is 0.5 atm?

Answer 1

To solve the problem, we can use the ideal gas law, which is given by the equation:

\[
PV = nRT
\]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (approximately 0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

Since the number of moles of gas does not change between the two states, we can write the relationship for the two conditions using \( P_1, V_1, T_1 \) for the initial conditions and \( P_2, V_2, T_2 \) for the final conditions:

\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]

Given:
- \( P_1 = 1 \, \text{atm} \)
- \( V_1 = 0.3 \, \text{L} \)
- \( T_1 = 300 \, \text{K} \)
- \( P_2 = 0.5 \, \text{atm} \)
- \( V_2 = 1 \, \text{L} \)

Now we can substitute these values into the equation:

\[
\frac{1 \times 0.3}{300} = \frac{0.5 \times 1}{T_2}
\]

Simplifying the left side:

\[
\frac{0.3}{300} = \frac{0.5}{T_2}
\]

Cross-multiplying to solve for \( T_2 \):

\[
0.3 T_2 = 150
\]

\[
T_2 = \frac{150}{0.3} = 500 \, \text{K}
\]

Thus, the final temperature \( T_2 \) is **500 K**.