Last updated on February 24th, 2022 at 08:21 pm

In this post, we will solve a few Numerical Problems based on the **Combined Gas law (or Gas equation**) & the **Ideal gas equation**.

The Formulas used to solve the numerical problems are:

**PV****/ T = constant (k)***p*V =*n*RT

**Example 1**

*A sample of nitrogen occupies a volume of 1.0 L at a pressure of 0.5 bar at 40°C. Calculate the pressure* *if the gas is compressed to 0.225 mL at –6°C.*

**Solution: ***p*1 = 0.5 bar

*p*2 = ?

V1 = 1.0 L

V2 = 0.225 mL = 0.225 × 10^{–3} L

T1 = 40+273 = 313 K

T2 = – 6 + 273 = 267 K

According to combined gas law equation,

**p1V1/T1 = p2V2/T2**

p2 = (p1V1T2)/(T1V2) = (.5 x 1x 267)/(313x 0.225 × 10^{–3}) = **1895.6 bar**

**Example **2

*At 25°C and 760 mm of Hg pressure, a gas occupies* *600 mL volume. What will be its pressure at a* *height where temperature is 10°C and volume of* *the gas is 640 mL.*

**Solution: ***p*1=760 mm Hg

*p*2 = ?

V1=600 mL

V2 = 640 mL

T1=273 + 25 = 298 K

T2= 273 + 10 = 283 K

According to combined gas law equation,

**p1V1/T1 = p2V2/T2**

**p2 = (p1V1T2)/(T1V2) = [**(760 mm Hg) (600 mL) (283 K)] / [(298K)(640 mL) ] = 676.6 mm Hg

**Example **3

*Calculate the moles of hydrogen (H2) present in a 500 mL sample of hydrogen gas at a pressure of 1* *bar and 27°C.*

**Solution:**

**1 atm = 1.01325 bar = 1.01325 × 10**^{5}**Pa = 101.325 kPa**

or **1 bar = 0.987 atm**

According to ideal gas equation,

*p*V = *n*RT

*p *= 1 bar = 1 atm = 101 x 10^3 Pa

V = 500 mL = 500 cm^{3} = 500 × 10^{–}^{6} m^{3}

T = 27 + 273 = 300 K,

R = **8.314 J mol**^{–1}**K**^{–1}

Now, *n *= pV/(RT) = 101 x 10^3 x 500 × 10^{–}^{6} / ( **8.314x 300) = 0.02 mole**

**Example **4

*Calculate the volume occupied by 4.045 × 10*^{23 }*molecules of oxygen at 27°C and having a pressure* *of 0.933 bar.*

* Solution: *Here, number of molecules = 4.045 × 10

^{23}

*p *= 0.933 bar = 0.933 atm = 0.933 x 101 x 10^3 Pa

T = 27 + 273 = 300 K

R =**8.314 J mol**^{–1}**K**^{–1}

**W****e will use *** p*V =

*n*RT

**.**V = nRT/p

Let us first calculate the number of moles, n.We know that number of moles n = No. of molecules / *6.022 **x 10*^{23}

*= **4.045 × 10*^{23}/ *6.022 x 10*^{23} = 0.672 mol

V = nRT/p = 0.672 x 8.314 x 300 / ( 0.933 x 101 x 10^3) = 0.0177 m^3

**Example **5

*A discharge tube of 2 L capacity containing hydrogen gas was evacuated till the pressure inside is 1 × 10*^{–5}*atm. If the tube is maintained at a temperature of 27°C, calculate the number of hydrogen molecules still present in the tube.*

**Solution:**

**Step I. ***To calculate the number of moles of hydrogen.*

Here *p *= 1 × 10^{–5} atm = 1 × 10^{–5 }x 101 x 10^{3} Pa = 101 x 10^{-2} Pa

V = 2 litres =2 000 cm^3 = 2000 x 10^{-6} m^{3} = 2 x 10^{-3} m^{3}

R = **8.314 J mol**^{–1}**K**^{–1}

T = 27 + 273 = 300 K

*n *= ?

According to the gas equation,

*p*V = *n*RT

or *n *= pV/RT = 101 x 10^{-2} x 2 x 10^{-3}/ (8.314 x 300) = 8.1 x 10^{-7} mole

**Step II. ***To calculate the number of hydrogen molecules.*

We know that

1 mol of hydrogen = 6.022 × 10^{23 }molecules

8.1 x 10^{-7} mole = 6.022 × 10^{23 }x 8.1 x 10^{-7 }= 48.77 x 10^{16} molecules

**Example **6

*Calculate the mass of 120 mL of N*_{2}* at 150°C and 1 × 10*^{5 }*Pa pressure.*

**Solution: **According to ideal gas equation,

*p*V = *n*RT =(*m/*M)RT

m = pVM/(RT)

*p *= 10^{5} Pa , V = 120 ×10^{–6} m^{3}

M = 28, T = 273 + 150 = 423 K,

R = 8.314 Nm K^{–1} mol^{–1}

Mass of gas m = pVM/(RT) =(10^{5} x 120 × 10^{–6 }x 28)/(8.314 x 423) = **0.0955 g**