Part A For each exercise, you can simulate the described conditions by changing

Question

Part A

For each exercise, you can simulate the described conditions by changing the values in the Run Experiment tool of the Simulation. To be able to measure the effects on pressure or volume, slide the Rspd bar in the Properties box to either P (atm) or V (L), respectively. The slider bar for either pressure or volume will turn yellow when it is selected, and it becomes the only dependent variable, i.e., the value you measure in response to changing the other properties. Note that when either pressure or volume is selected with Rspd, the respective value can no longer be directly controlled.

Let us first examine the behavior of an ideal gas when we force the volume to be a value of our choosing. We can examine how changes to the absolute temperature and number of moles affects the pressure of the gas particles (by selecting pressure with Rspd such that pressure cannot be controlled).

There are 0.05 mol of neon at a temperature of 265.00 K occupying a volume of 1.40 L . Use the Run Experiment tool in the Simulation to determine the initial pressure of the gas and the new pressure values when the temperature or number of moles is changed to complete the following statements.

Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.

Part B

For this exercise, you can simulate the described conditions by changing the values in the Run Experiment tool of the Simulation. To be able to measure the effects on the gas volume, slide the Rspd bar in the Properties box to V (L) so that the volume bar turns yellow. This way, volume becomes the only dependent variable. Note that when volume is selected with Rspd, its value can no longer be directly controlled.

Suppose a piston automatically adjusts to maintain a gas at a constant pressure of 8.20 atm . For the initial conditions, there are 0.05 mol of argon at a temperature of 255.00 K . This gas occupies a volume of 0.13 L under those conditions. What volume will the gas occupy if the number of moles is increased to 0.08 mol (n2) from the initial conditions? What volume will the gas occupy if the temperature is increased to 355.00 K (T2) from the initial conditions?

Express the volumes in liters to two decimal places separated by a comma.

View Available Hint(s)

volume at n2, volume at T2 =

Reset Help 1.40 atm The initial pressure of the neon gas is 0.78 atm From the initial conditions (0.05 mol and 265.00 K), if you increased the amount of neon to 0.09 mol, the new pressure would be number of moles in 0.44 atm Therefore, pressure with the 1.55 atm proportion. direct From the initial conditions (0.05 mol and 265.00 K), if you increased the temperature by 80.00 K, the new pressure would be temperature in decreases Therefore, pressure with proportion. increases 0.87 atm 1.82 atm 1.01 atm inverse

Explanation / Answer

Part A

Use the ideal gas law: P*V = n*R*T where P = pressure of the gas, V = volume of the gas, n = number of moles of the gas and T = temperature of the gas.

We have n = 0.05 mole, V = 1.40 L and T = 265.00 K. Plug in values and obtain

P = n*R*T/V = (0.05 mole)*(0.082 L-atm/mol.K)*(265.00 K)/(1.40 L) = 0.776 atm 0.78 atm (ans).

The volume and temperature remain the same; however, n = 0.09 mole now. We can easily see that V and T remaining constant, P1/P2 = n1/n2. Put P1 = 0.78 atm, n1 = 0.05 mole and n2 = 0.09 mole and obtain

P2 = P1*n2/n1 = (0.78 atm)*(0.09 mole)/(0.05 mole) = 1.404 atm 1.40 atm (ans).

Therefore, pressure increases with the number of moles in direct proportion.

The temperature is increased by 80.00 K so that the new temperature is (265.00 + 80.00) K = 345.00 K. We can easily see that V and n remaining constant, P1/P2 = T1/T2 or

P2 = P1*T2/T1 = (0.78 atm)*(345.00 K)/(265.00 K) = 1.015 atm 1.01 atm (ans).

Therefore, pressure increases with temperature in direct proportion.

Part B

Again, use the ideal gas law as before. We have n1 = 0.05 mole, V1 = 0.13 L and n2 = 0.08 mole. It is not difficult to easily deduce that

V1/V2 = n1/n2 where V2 is the volume when n2 = 0.08 mole. Plug in values and obtain

V2 = V1*n2/n1 = (0.13 L)*(0.08 mole)/(0.05 mole) = 0.208 L (ans).

Again, we have T2 = 355.00 K and T1 = 255.00 K; it is easy to deduce V1/V2 = T1/T2. Plug in values and obtain

V2 = V1*T2/T1 = (0.13 L)*(355.00 K)/(255.00 K) = 0.181 L (ans).

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