Answer:
[tex]\boxed {\boxed {\sf 232.9 \textdegree C}}[/tex]
Explanation:
This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:
[tex]\frac {V_1}{T_1}= \frac{V_2}{T_2}[/tex]
The volume of the gas starts at 250 milliliters and the temperature is 137 °C.
[tex]\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}[/tex]
The volume of the gas is increased to 425 milliliters, but the temperature is unknown.
[tex]\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}[/tex]
We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.
[tex]250 \ mL * T_2 = 137 \textdegree C * 425 \ mL[/tex]
Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.
[tex]\frac{250 \ mL * T_2}{250 \ mL}=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}[/tex]
[tex]T_2=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}[/tex]
The units of milliliters (mL) cancel.
[tex]T_2=\frac{ 137 \textdegree C * 425 }{250 }[/tex]
[tex]T_2= \frac{58225}{250} \textdegree C[/tex]
[tex]T_2=232.9 \textdegree C[/tex]
The temperature changes to 232.9 degrees Celsius.
