Question

If the temperature of a gas is increased from 20°C to 35°C, what is the new pressure if the original pressure was 1.2 atm? Assume that volume is constant.Group of answer choicesA. 0.6 atmB. 1.3 atmC. 1.0 atmD. 2.6 atm

Answer 1

ANSWER

The new pressure of the gas is 1.3atm

EXPLANATION

Given that;

The initial temperature of the gas is 20 degrees celcius

The final temperature of the gas is 35 degrees celcius

The initial pressure of the gas is 1.2 atm

Follow the steps below to find the final pressure of the gas

Step 1; Convert the temperature to degrees Kelvin using the below conversion formula

$\text{ T K = t}\degree C\text{ + 273.15}$$\begin{gathered} \text{ For T1 = 20}\degree C \\ \text{ T = 20 + 273.15} \\ \text{ T = 293.15K} \\ \\ \text{ For T}_2\text{ = 35}\degree C \\ \text{ T}_2\text{ = 35 + 273.15} \\ \text{ T}_2\text{ = 308.15K} \end{gathered}$

Step 2; Apply the Gay Lussac's law

$\text{ }\frac{\text{ P}_1}{\text{ T}_1}\text{ = }\frac{\text{ P}_2}{\text{ T}_2}$

Substitute the given data into the above formula

$\begin{gathered} \text{ }\frac{\text{ 1.2}}{\text{ 293.15}}\text{ = }\frac{\text{ P}_2}{\text{ 308.15}} \\ \text{ cross multiply} \\ \text{ 1.2 }\times\text{ 308.15 = P}_2\text{ }\times\text{ 293.15} \\ \text{ 369.78 = 293.15 P}_2 \\ \text{ Divide both sides by 293.15} \\ \text{ P}_2\text{ = }\frac{\text{ 369.78}}{\text{ 293.15}} \\ \text{ P}_2\text{ = 1.26 atm} \\ P_2\approx\text{ 1.3 atm} \end{gathered}$

Therefore, the new pressure of the gas is 1.3 atm