Down below is the question... Can you please show work and answer thanks!

Question

At 134 degrees celsius, the pressure of a sample of nitrogen is 1.15 atm. What will the pressure be at 243 degrees celsius, assuming constant volume? 

Answer 1

Hi Joanna.

 

The best way to approach these problems is with the ideal gas law:

 

PV = nRT

 

P = Pressure

V = Volume

n = number of moles of your gas

R = gas constant

T = temperature in K

 

Part 1:

 

You want to set up equations to solve for your unknown pressure. We know from the problem that the volume remains constant. Since we aren't inserting anymore nitrogen into our system, the number of moles of nitrogen also remain constant, and of course R remains constant. Your only variables are pressure and temperature.

 

Be careful with this problem though; it reports the temperature in deg C. But with ideal gas law, the temperature needs to be K.

 

K₁ = 134 + 273.15 = 407.15 K

K₂ = 243 + 273.15 = 516.15 K

 

P₁V = nRT₁

P₂V = nRT₂

 

From equation 1, solve for n:

 

n = P₁V/(RT₁)

 

Plug it into equation 2.

 

P₂V = (P₁V/(RT₁))*RT₂

 

Cancel out the R's and V's

 

P₂ = P₁T₂/T₁

 

P₂ = (1.15atm)(516.15)/(407.15)

 

= 1.46 atm

 

**

 

Part 2:

 

You initially have a gas occupying 22mL at standard conditions. Standard conditions means 15 deg C and 1atm. You know from chemistry that 1 mol of a gas occupies 22.4 L. So your n value will be:

 

n = (22mL * 1L / 1000mL) / (22.4 L / mol) = 9.8 x 10^-4 moles.

 

It then undergoes a rise in temperature to 16 deg C (289.15 K) and a reduction in pressure to 0.9241 atm.

 

Using the same equation above:

 

PV = nRT

 

P = 0.9241 atm

V = ???

n = 9.8 x 10^-4 moles

R = 0.082 L atm mol^-1 K^-1

T = 289.15 K

 

Note: choosing the correct value for R is essential in solving these equations. Because pressure, volume, and temperature can be reported with so many different units, you can R values of 8.314, 1.987, 82.055, etc. You need to make sure all your units match up with it. Here, you're reporting pressure in atm, volume in L, and temperature in K. That's why the R value has all those units.

 

Now that you have all your values assembled, you can solve for V.

 

V = (9.8 x 10^-4)*(0.082)*(289.15)/(0.9241)

 

V = 0.0251 L

 

or

 

V = 25.1 mL

 

Hope this helps.