Question

A sample of helium has a temperature of 450 K. The gas is cooled to 248.9 K at which time the gas occupies 103.4 L? Assume pressure is constant at 3 atm. What was the original volume of the gas?

A. 186.94 L
B. 304.5 L
C. 57.19 L
D. 361.07 L

Answer 1

Answer:

$\boxed {\boxed {\sf A. \ 186.94 \ L}}$

Explanation:

We are asked to find the original volume of a gas given a change in temperature. Since pressure remains constant, we are only concerned with volume and temperature, so we use Charles's Law. This 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 gas begins with a temperature of 450 Kelvin, but the volume is unknown.

$\frac {V_1}{450 \ K }= \frac{V_2}{T_2}$

The gas is cooled to 248.9 Kelvin and the gas occupies a volume of 103.4 liters.

$\frac {V_1}{450 \ K }= \frac{103.4 \ L}{248.9 \ K}$

Since we are solving for the original volume, we must isolate the variable V₁. It is being divided by 450 Kelvin. The inverse operation of division is multiplication, so we multiply both sides of the equation by 450 K.

$450 \ K \frac {V_1}{450 \ K }= \frac{103.4 \ L}{248.9 \ K}* 450 \ K$

$V_1= \frac{103.4 \ L}{248.9 \ K}* 450 \ K$

The units of Kelvin cancel.

$V_1= \frac{103.4 \ L}{248.9 }* 450$

$V_1= 0.4154278827 \ L *450$

$V_1= 186.9425472 \ L$

Round to the nearest hundredth. The 2 in the thousandths place tells us to leave the 4 in the hundredth place.

$V_1 \approx 186.94 \ L$

The original volume is approximately 186.94 liters and Choice A is correct.