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?

1 answer

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.