Question

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

Answer 1

Question

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

Answer

$= \sf 676.633 \: mm \: of \: Hg$

Solution

Let's proceed with observing the given data :-

$\boxed{ \begin{array}{c|c} \sf P_1 = 760 \: mm& \sf \: P_2 = ? \ \\ \sf V_1 = 600 \: ml & \sf \: V_2 =640 \:ml\\ \sf \: T_1 = 25 \degree C& \sf T_2 = 10 \degree \: C \end{array}}$

About

It's given that at 25° C optimising temperature and with pressure 760 mm of Hg , a gas , occupies a volume of 600 ml . Now , if temperature reduces to 10° C at a certain pressure , the gas occupies a volume of 640 ml .

Thus , we have to calculate that final pressure.

Here ,

Initial temperature = 25° C = 25 + 273 = 298 K

Final temperature = 10° C = 10 + 273 = 283 K

On apply Gas Law ,

$\sf \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$

$↦\sf \dfrac{760 \times 600}{298} = \dfrac{P_2 \times 640}{283}$

$↦ \sf \: P_2 = \dfrac{760 \times 600 \times 283}{298 \times 640}$

$↦ \sf \: P_2 = \dfrac{129048 \times 1000}{19072 \times 10}$

$↦ \sf \: P_2 = 676.633 \: mm \: of \: Hg$

Conclusion

When temperature would reduce to 10° C and pressure of 676.633 mm is supplied the gas would occupy the volume of 640 ml .

$\rule{200pts}{2pt}$

Gas Law

This law states that the ratio of the product of volume and pressure and the gas’s absolute temperature is equal to a constant .The combined gas law combines is a law that combines the three gas laws i.e Boyle’s Law , Charles Law, and Gay-Lussac’s Law .

At STP ( Standard Temperature and Pressure )

T = 273 K = 0° C

P = 760 mm of Hg = 76 cm of Hg = 1 atm

V = 22.4 L ( 1 mole of ideal gas )

Ideal Gas Equation

PV = nRT

n = number of moles

R = Universal gas Constant

R = 0.0821 Latm/mol K

Thankyou