Question

A gas has an initial volume of 52.3 L at 273 Kelvin. What is its temperature when the volume reaches 145.7 L?

Answer 1

Answer:

$\boxed {\boxed {\sf 761 \ K}}$

Explanation:

We are asked to find the new temperature of a gas after a change in volume. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The volume is initially 52.3 liters at a temperature of 273 Kelvin.

$\frac {52.3 \ L}{273 \ K}= \frac{V_2}{T_2}$

The volume reaches 145.7 liters at an unknown temperature.

$\frac {52.3 \ L}{273 \ K}= \frac{145.7 \ L }{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂.  Cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

$52.3 \ L * T_2 = 273 \ K * 145.7 \ L$

Now the variable is being multiplied by 52.3 liters. The inverse of multiplication is division. Divide both sides by 52.3 L.

$\frac {52.3 \ L * T_2 }{52.3 \ L}=\frac{ 273 \ K * 145.7 \ L}{52.3 \ L}$

$T_2=\frac{ 273 \ K * 145.7 \ L}{52.3 \ L}$

The units of liters cancel.

$T_2=\frac{ 273 \ K * 145.7 }{52.3 }$

$T_2 = 760.5372849 \ K$

The original measurements have at least 3 significant figures, so our answer must have 3. For the number we found, that is the ones place. The 5 in the tenths place tells us to round the 0 up to a 1.

$T_2 \approx 761 \ K$

When the volume reaches 145.7 liters, the temperature is 761 Kelvin.