Question

A 7.70 L container holds a mixture of two gases at 51 ° C. The partial pressures of gas A and gas B, respectively, are 0.177 atm and 0.726 atm. If 0.210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

0.12

Step-by-step explanation:

Di the test

Answer 2

`P_(total)=1.628atm`

Step-by-step explanation:

According to dalton theory of partial pressure , partial pressure applied by a component in a mixture of more gas is same as pressure applied when it is taken alone in the container.

And thus
`P_(total)=P_A+P_B+P_C`

and P_A and P_B is given so we need to find P_C for the 3rd gas,

and we will solve by assuming that gas follows the ideality nature of gas,

PV=nRT

P_A=0.177atm

P_B=0.726atm

`P_C * 7.7L =0.21 * 0.0821 * 324`

`P_C=0.725atm`

therefore,

`P_(total)=P_A+P_B+P_C`

`P_(total)=0.177atm +0.726atm + 0.725atm`

`P_(total)=1.628atm`