Question

A gas has a volume of 4.25 m3 at a temperature of 95.0°C and a pressure of 1.05 atm. What temperature will the gas have at a pressure of 1.58 atm and a volume of 2.46 m3

Answer 1

Answer:

$\boxed {\boxed {\sf 82.7 \textdegree C}}$

Explanation:

We are asked to find the temperature of a gas given a change in pressure and volume. We will use the Combined Gas Law, which combines 3 gas laws: Boyle's, Charles's, and Gay-Lussac's.

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

Initially, the gas has a pressure of 1.05 atmospheres, a volume of 4.25 cubic meters, and a temperature of 95.0 degrees Celsius.

$\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{P_2V_2}{T_2}$

Then, the pressure increases to 1.58 atmospheres and the volume decreases to 2.46 cubic meters.

$\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{1.58 \ atm *2.46 \ m^3}{T_2}$

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

$(1.05 \ atm * 4.25 \ m^3) * T_2 = (95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)$

Now the variable is being multiplied by (1.05 atm * 4.25 m³). The inverse operation of multiplication is division, so we divide both sides by this value.

$\frac {(1.05 \ atm * 4.25 \ m^3) * T_2}{(1.05 \ atm * 4.25 \ m^3)} = \frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}$

$T_2=\frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}$

The units of atmospheres and cubic meters cancel.

$T_2=\frac{(95.0 \textdegree C)*(1.58* 2.46 )}{(1.05 * 4.25 )}$

Solve inside the parentheses.

$T_2= \frac{(95.0 \textdegree C)*3.8868}{4.4625}$

$T_2= \frac{369.246}{4.4625} \textdegree C}$

$T_2 = 82.74420168 \textdegree C$

The original values of volume, temperature, and pressure all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 4 in the hundredth place to the right tells us to leave the 7 in the tenths place.

$T_2 \approx 82.7 \textdegree C$

The temperature is approximately 82.7 degrees Celsius.