Question

a gas is contained in a thick - walled balloon with an initial temperature of 303 k. when the pressure changes from 100.0 kpa to 90.0 kpa, the volume changes from 2.50 l to 3.75 l. what is the final temperature of the system?

Explanation:

Step1: Use combined gas law

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

Step2: Rearrange for final temperature

$$T_2 = \frac{P_2V_2T_1}{P_1V_1}$$

Step3: Substitute values

$$T_2 = \frac{(90.0\ \text{kPa})(3.75\ \text{L})(303\ \text{K})}{(100.0\ \text{kPa})(2.50\ \text{L})}$$

Step4: Calculate result

$$T_2 = \frac{90.0 \times 3.75 \times 303}{100.0 \times 2.50} = 409.05\ \text{K}$$

Answer:

409 K (or 409.05 K)