Looking in general, we will find that gases are made up of very small particles, and due to the mutual collisions between these particles, they keep moving continuously, due to which their nature is elastic, during the interaction of their elastic action or collision between them makes them an ideal gas. This tolerance and elastic nature are different in all gases, depending on their nature and properties. Some such gases are nitrogen, oxygen, hydrogen, carbon dioxide, noble gases, etc., which can be divided as an ideal gas.

PV = nRT

where,

P is the pressure

V is the volume

n is the amount of substance

R is the ideal gas constant

Ideal gas mainly follows some basic rules like Boyle’s law, Charles / Gelusac law, related to which an expression has also been constructed, which is related to an ideal gas, we find it as PV = nRT. Here the value of R describes the universal gas constant with a value of 8.31 J/mol degree and K. Also, in SI units, pressure is expressed in N/m2 (Pascals), volume is expressed in cubic metres and temperature in degrees Kelvin.

Gas Law Problems

Question: At ordinary temperature, the gas takes up 4.00 L. What is the volume at 440.0 degrees Celsius?

Solution:

In cross-multiplied form, it is this:

V1T2 = V2T1

V2 = (V1) [T2 / T1] <— notice how I grouped the temperatures together

x = (4.00 L) [(717.0 K) / (273.0 K)]

x = 10.50 L

Question: The temperature of a gas will change from 405 °C to 552 °C when the volume is changed from mL to 850 mL. What’s the starting point for the volume?

Solution:

Write Charles Law and substitute values in:

V1 / T1 = V2 / T2

y / 678 K = 850 mL / 725 K

(y) (725 K) = (852 mL) (678 K)

y = 850 mL

The huge °C figures are intended to make you forget to add 273. Keep in mind that only Kelvin temperatures can be used in the calculations.

Question: When the temperature of 500 ml of air rises from 23°C to 43°C under constant pressure, what is the increase in volume?

Solution:

Charle’s law is applicable as the pressure and amount remain constant.  

V1 / T1 = V2 / T1 or V­1 = T1 / T2 × V2  

V1/500 = 316 / 296  = 533.783ml  

 Increase in volume of air = 533.783 – 500 = 33.783 ml 

Question: Under atmospheric pressure, 2 litres of gas is equal to two-litre of liquid. What is the volume of the same amount of gas at the same temperature under 550 mm of Hg?

Solution:

Given V1 = 2 litre

P1 = 2 atm

V2 = ? 

P2 = 550 / 760 atm

Using

P1V1 = P2V2

We get

2 × 2 = 550 / 760 × V2

V2 =760 litre

Question: At 24°C and 760 mm of Hg, how big of a balloon could you fill with 6g of He gas?

Solution:               

Given, P = 760 / 760 atm, T = 297K, w = 6g  

and m = 4 for He  

PV = w / M RT  

 = 760 / 760 × V = 6/4 × 1 × 297  

∴ V = 445.5 litre    

Question: Calculate the temperature at which 32g N2 occupies a volume of 8 litres at 

4.46 atm    

Solution:     

w = 32g, P = 4.46 atm, V = 8 litre, m = 28  

Now, PV = w / M RT (R = 0.0821 litre atm K–1 mole–1)  

T = 2.563 K 

Question: What is the volume of a gas that occupies 400 ml at 37°C and 640 mm pressure?

Solution:            

V2 = 400 / 1000 litre, P2 = 640 / 760 atm, T2 = 310K  

At STP, V1= ? P1= 1 atm, T1= 273K  

P2V2 / T2 = P1V1 / T1 or V1 = 0.8258 litre  

Volume at STP = 825.8 ml

Question: A 42-cm-diameter cylindrical balloon will be filled with H2 at NTP from a cylinder storing the gas at 10 atm and 37°C. At NTP, the cylinder can contain 3.82 litres of water. Determine the maximum number of balloons that can be filled.

Solution:

Volume of 1 balloon which has to be filled = 4/3 π (42/2)3 =9.702 litre  

Let x balloons be filled, then volume of H2 occupied by balloons = 9.702 × x  

Also, the cylinder will not be empty and it will occupy a volume of H2 = 3.82 litre.  

∴ Total volume occupied by H2 at NTP = 9.702 × x + 3.82 litre  

∴ At STP  

P2 = 1 atm   Available H2  

V1= 9.702 × x + 3.82        P2 = 10 atm  

T1 = 273 K   T2 = 310K  

P1V1 / T2 = P2V2 / T2  V2 =  3.82 litre  

or 1 × (4.85 1x + 3.82 / 273) = 10 × 3.82 / 310 ∴ x = 7 

Question: What would be the final pressure in an empty 0.45 litre wire bottle after all CO2 has been dissipated and the temperature reaches 35°C if a 40g chunk of dry ice is placed in it and tightly closed?

Solution:         

w = 40g dry CO2 which will evaporate to develop pressure  

m = 44, V = 0.45 litre, P = ? T = 308K  

PV = W / m RT  

P × 0.45 = 40 / 44  × 0.0821 × 308  

P = 0.07989 atm  

Pressure inside the bottle = P + atm pressure  = 0.07989+ 1 = 1.07989 atm

Conclusion

The ideal gas law does not apply because the molecular size and intermolecular forces matter at low temperatures, large densities, and extremely high pressures. The ideal gas law does not apply. For heavy gases (refrigerants) and gases with significant intermolecular interactions, such as Water Vapour, the ideal gas law does not apply.