Question

combined gas law

1. a gas has a volume of 2.5 l at 300k and 1.0 atm. what will the volume be at 2.0 atm and 400 k?

Explanation:

Step1: Identify the Gas Law

This problem involves pressure and volume changes of a gas, so we use Boyle's Law (for constant temperature, \( P_1V_1 = P_2V_2 \)). Wait, but there's also a temperature change? Wait, the initial temperature is 300K, final temperature? Wait, the problem says: "A gas has a volume of 2.5L at 300K and 1.0 atm. What will the volume be at 2.0 atm and 400K?" Wait, let's parse the problem correctly. Initial conditions: \( P_1 = 1.0 \, \text{atm} \), \( V_1 = 2.5 \, \text{L} \), \( T_1 = 300 \, \text{K} \). Final conditions: \( P_2 = 2.0 \, \text{atm} \), \( T_2 = 400 \, \text{K} \), find \( V_2 \). So we use the Combined Gas Law: \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \).

Step2: Rearrange the Combined Gas Law

Solve for \( V_2 \): \( V_2 = \frac{P_1V_1T_2}{P_2T_1} \)

Step3: Substitute the Values

\( P_1 = 1.0 \, \text{atm} \), \( V_1 = 2.5 \, \text{L} \), \( T_1 = 300 \, \text{K} \), \( P_2 = 2.0 \, \text{atm} \), \( T_2 = 400 \, \text{K} \)

\( V_2 = \frac{(1.0 \, \text{atm})(2.5 \, \text{L})(400 \, \text{K})}{(2.0 \, \text{atm})(300 \, \text{K})} \)

Step4: Calculate the Numerator and Denominator

Numerator: \( 1.0 \times 2.5 \times 400 = 1000 \)

Denominator: \( 2.0 \times 300 = 600 \)

Step5: Divide to Find \( V_2 \)

\( V_2 = \frac{1000}{600} \approx 1.67 \, \text{L} \)

Answer:

The volume will be approximately \( 1.67 \, \text{L} \) (or \( \frac{5}{3} \, \text{L} \) if exact fraction).