Question

1) A sample of an ideal gas has a volume of 2.27 L at 285 K and 1.10 atm. Calculate the pressure when the volume is 1.36 L and the temperature is 306 K.

2) A 7.90 L container holds a mixture of two gases at 51 °C. The partial pressures of gas A and gas B, respectively, are 0.292 atm and 0.676 atm. If 0.110 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

Answer:

Question 1 ; 2.15 atm

Question 2; 1.078 atm

Step by step explanation:

Question 1

We can use the Ideal Gas Law to calculate the pressure: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant (8.31 J/mol·K), and T is the temperature in Kelvin.

Given the initial conditions:

V1 = 2.27 L, T1 = 285 K, and P1 = 1.10 atm

And the final conditions:

V2 = 1.36 L, T2 = 306 K

We can rearrange the Ideal Gas Law to solve for P2:

P2 = (nRT2) / V2 = (nRT1) / V1 * (V1 / V2) = (P1 * V1 * T2) / (T1 * V2) = (1.10 atm * 2.27 L * 306 K) / (285 K * 1.36 L) = 2.15 atm

Question 2

The total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases. So, after adding the third gas, the total pressure will become:

0.292 atm + 0.676 atm + 0.110 atm = 1.078 atm.