Question

A sample of nitrogen gas occupies a volume of 2.55 L when it is at 755 mm Hg and 23 degrees Celsius. Use this information to determine the moles of nitrogen gas present. Volume is increased to 4.10 L and temperature is decreased to 18 degrees Celsius. Calculate the final pressure

Answer 1

Answer:

The final pressure is 461.6 mm Hg

Explanation:

Step 1:Data given

Volume of N2 = 2.55L

Pressure N2 = 755 mm Hg

Temperature = 23.0 °C

New volume = 4.10 L

Temperature = 18.0 °C

Step 2: Calculate the fina pressure

P1V1/T1 = P2V2/T2

P2 = P1V1/T1 * T2/V2

⇒ with P1 = initial pressure = 755/760 atm =0.993421 atm

⇒ with V1 = the initial volume = 2.55L

⇒ with T1 = the initial temperature = 23.0 °C = 296 K

⇒ with P2 = The final pressure

⇒with V2 = the new volume = 4.10 L

⇒with T2 = the new temperature = 18.0 °C = 291 K

P2 = 755 mm *2.55 L *296K/(291 K* 4.10 L) = 461.6 mm Hg

The final pressure is 461.6 mm Hg

Answer 2

Answer: The number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg

Explanation:

To calculate the amount of nitrogen gas, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 755 mmHg

V = Volume of the gas = 2.55 L

T = Temperature of the gas = $23^oC=[23+273]K=296K$

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$755mmHg\times 2.55L=n\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 296K\\\\n=\frac{755\times 2.55}{62.364\times 296}=0.1043mol$

To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=755mmHg\\V_1=2.55mL\\T_1=23^oC=[23+273]K=296K\\P_2=?\\V_2=4.10L\\T_2=18^oC=[18+273]K=291K$

Putting values in above equation, we get:

$\frac{755mmHg\times 2.55L}{296K}=\frac{P_2\times 4.10L}{291K}\\\\P_2=\frac{755\times 2.55\times 291}{4.10\times 296}=461.6mmHg$

Hence, the number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg