Login
Get Advice on Live Video Call
Earn $ Cash $ with
consultations on Bissoy App

You have a sample of neon gas at a certain pressure, volume, and temperature. You double the volume, double the number of moles of neon, and double the Kelvin temperature. How does the final pressure (Pf) compare to the original pressure (Po) ?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

The final pressure will double as well.

If you use the equation, $PV = nRT$, to express the initial and the final states of your gas sample, you can write something like this

$P_1V_1 = n_1 * R * T_1$ $->$ the initial state; (1)
$P_2V_2 = n_2 * R * T_2$ $->$ the final state;

You also know that the volume doubles, which means that $V_2 = 2 * V_1$, the number of moles doubles, $n_2 = 2 * n_1$, and the temperature doubles, $T_2 = 2 * T_1$. Use these equalities into the equation that describes the final state

$P_2 * 2 * V_1 = 2 * n_1 * R * 2 * T_1$ (2)

Divide equation (1) by equation (2) and you'll get

$P_1/(2 * P_2) = (n_1 * R * T_1)/(2 * n_1 * R * 2 * T_1)$

$P_1/( 2 * P_2) = 1/4 => 2 * P_2 = 4 * P_1 => P_2 = 2 * P_1$

Therefore, the pressure doubles as well. Finally, the answer makes sense because the temperature and the number of moles will each double the pressure, but the fact that the volume doubles as well will cancel one of these effects out.

If the volume would have stayed the same, the pressure would have increased 4 times - 2 times because the temperature doubled and 2 times more because the number of moles doubled.

0 like 0 dislike
1 views
Under certain conditions of temperature and pressure, hydrogen gas and nitrogen gas react to form ammonia gas. Determine the mass of ammonia produced if 3.50 moles of hydrogen has reacts with 5.00 moles of nitrogen gas?

Start by balancing the equation for the reaction between hydrogen $"H"_2$ and nitrogen $"N"_2$, which produces ammonia $"NH"_3$: $color(darkgreen)(1) color(white)(.) "N"_2 (g) + color(darkgreen)(3) color(white)(.) "H"_2 to color(darkgreen)(2) color(white)(.) "NH"_3...

1 Answers 1 views
If a gas in a closed container is pressurized from 15.0 atm to 16.0 atm and its original temperature was 25°C, what would be the final temperature of the gas in Kelvin?

Gay-Lussac's Law deals with pressure and temperature. $P_1/T_1 = P_2/T_2$ If we rearrange and get $T_2$ by itself, we get $T_2= (P_2T_1)/ P_1$ Since temperature needs to be in Kelvin...

1 Answers 1 views
Is gas volume of 444 mL at 273K and 79.0 kPa. What is the final kelvin temperature when the volume of the gas is changed to 1880 mL and the pressure changed to 38.7 kPa?

We're asked to calculate the final absolute (Kelvin) temperature of a gas system when it is subdued to known pressure and volume changes. To solve this problem, we can use...

1 Answers 1 views
A 3.50-L gas sample at 20°C and a pressure of 86.7 kPa expands to a volume of 8.00 L. The final pressure of the gas is 56.7 kPa. What is the final temperature of the gas, in degrees Celsius?

We can use here the , relating the temperature, pressure, and volume of a gas with a constant quantity: $(P_1V_1)/(T_1) = (P_2V_2)/(T_2)$ If you're using the ideal-gas equation, which we're...

1 Answers 1 views
What is the initial pressure of a gas having an initial temperature of 90.5 K, an initial volume of 40.3 L, a final pressure of 0.83 atm, a final temperature of 0.54 K and a final volume of 2.7 L?

$PV = nRT$ $(PV)/T = nR$ $therefore (P_1V_1)/T_1 = (P_2V_2)/T_2$, the . $(P_1*"40.3 L")/("90.5 K")=("0.83 atm"*"2.7 L")/("0.54 K")$ $P_1 approx 9.3atm$

1 Answers 1 views
A gas occupies a volume of 444 mL at 273 K and 79.0 kPa. What is the final kelvin temperature when the volume of the gas is change 880 mL and the pressure is changed to 38.7 kPa?

Using the you can solve for your missing variable. $(P1V1)/(T1)=(P2V2)/(T2)$ P1 = 79.0 kPa V1 = 444 mL T1 = 273 K P2 = 38.7 kPa V2 = 880...

1 Answers 1 views
The volume of a sample of a gas at 273 Kelvin and is 100.0 L. If the volume is decreased to 50.0 L at constant pressure, what will be the new temperature of the gas?

This is an example of , which states that the volume of a given amount of a gas is directly proportional to its Kelvin temperature. The formula to use for...

1 Answers 1 views
The pressure of a sample of dry air is held constant at 2.25 atm while the temperature is decreased rom 100 °C to 7 °C. The original volume of the sample is 43 L. What is the final volume of the sample?

We use old .......$VpropT$ at constant pressure. $(V_1)/T_1=(V_2)/T_2$ And solve for $V_2=(V_1)/(T_1)xxT_2$ $(43*L)/(373*K)xx280*K=??L$

1 Answers 1 views
What is the initial pressure of a gas having an initial temperature of 90.5K, an initial volume of 40.3 L, final pressure of .83 atm, a final temperature of .54 K and a final volume of 2.7L?

Use the . $(P_1V_1)/T_1=(P_2V_2)/T_2$ Given $V_1="40.3 L"$ $T_1="90.5 K"$ $P_2="0.83 atm"$ $V_2="2.7 L"$ $T_2="0.54 K"$ Unknown $P_1$ Equation $(P_1V_1)/T_1=(P_2V_2)/T_2$ Solution Rearrange the equation to isolate $P_1$ and solve. $P_1=(P_2V_2T_1)/(T_2V_1)$ $P_1=((0.83"atm"...

1 Answers 1 views
A 3.00 liter sample of gas is at 288 K and 1.00 atm. If the pressure of the gas is increased to 2.00 atm and its volume is decreased to 1.50 liters, what will be the Kelvin temperature of the sample?

When you are given this much information in the context of a generic, unnamed gas, it's a good idea to consider the : $PV = nRT$ Before we...

1 Answers 1 views

Log in to ask questions, provide answers, or leave comments.


Login

Install the Bissoy app to consult with a doctor.


Download