A flexible container is charged with 79.00 L of gas at 352 Κ. Under conditions of constant pressure and moles of gas, what is the volume of the gas when the temperature is decreased by a factor of three?

When the temperature is decreased by a factor of three, the volume will decrease as well by a factor of 3. In this situation, the new volume will be 26.33 L

Explanation:

Step 1: Data given

Volume of the container = 79.00 L

Temperature in the container = 352 K

The pressure and number of moles is constant.

The temperature is decreased by a factor of three

Step 2: Calculate the volume

V1/T1 = V2/T2

⇒with V1 = the initial volume = 79.00L

⇒with T1 = the initial temperature = 352 K

⇒with V2 = the final volume = TO BE DETERMINED

⇒with T2 = the final volume = T1/3 = 117.3 L

V2 = ((1/3 * T1) / V1) *T2

V2 = V1/3

79.00 / 352  = V2 / 117.3

V2 = 26.33

When the temperature is decreased by a factor of three, the volume will decrease as well by a factor of 3. In this situation, the new volume will be 26.33 L

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol