To find the temperature at which the volume of a gas is 1 L at a pressure of 0.5 atm, we can use the Ideal Gas Law, represented by the equation:
\[ PV = nRT \]
However, since we have two different states of the gas, we can use the combined gas law, which is given by:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \( P_1 \) = initial pressure = 1 atm
- \( V_1 \) = initial volume = 0.3 L
- \( T_1 \) = initial temperature = 300 K
- \( P_2 \) = final pressure = 0.5 atm
- \( V_2 \) = final volume = 1 L
- \( T_2 \) = final temperature (what we need to find)
Now, substituting in the known values:
\[ \frac{(1 , \text{atm})(0.3 , \text{L})}{300 , \text{K}} = \frac{(0.5 , \text{atm})(1 , \text{L})}{T_2} \]
Cross-multiplying to solve for \( T_2 \):
\[ 1 , \text{atm} \times 0.3 , \text{L} \times T_2 = 0.5 , \text{atm} \times 1 , \text{L} \times 300 , \text{K} \]
\[ 0.3 T_2 = 0.5 \times 300 \]
\[ 0.3 T_2 = 150 \]
Now, divide both sides by 0.3 to isolate \( T_2 \):
\[ T_2 = \frac{150}{0.3} = 500 , \text{K} \]
So the final answer is:
d) 500 K