Questions: A gas is contained in a thick-walled balloon. When the pressure changes from 100 kPa to 90.0 kPa, the volume changes from 2.50 L to 3.75 L and the temperature changes from 303 K to K.

Solution Steps

The problem involves changes in pressure, volume, and temperature of a gas. The ideal gas law, \( PV = nRT \), can be used to relate these variables. However, since the number of moles \( n \) and the gas constant \( R \) are constant, we can use the combined gas law:

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

Assign the given values to the variables in the combined gas law:

• Initial pressure, \( P_1 = 100 \, \text{kPa} \)
• Final pressure, \( P_2 = 90.0 \, \text{kPa} \)
• Initial volume, \( V_1 = 2.50 \, \text{L} \)
• Final volume, \( V_2 = 3.75 \, \text{L} \)
• Initial temperature, \( T_1 = 303 \, \text{K} \)
• Final temperature, \( T_2 = ? \, \text{K} \)

Rearrange the combined gas law to solve for the final temperature \( T_2 \):

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

Substitute the known values into the equation:

\[ T_2 = \frac{90.0 \, \text{kPa} \times 3.75 \, \text{L} \times 303 \, \text{K}}{100 \, \text{kPa} \times 2.50 \, \text{L}} \]

Perform the calculation:

\[ T_2 = \frac{90.0 \times 3.75 \times 303}{100 \times 2.50} = \frac{101737.5}{250} = 406.95 \, \text{K} \]

Final Answer