Question

if a sample of gas at 25.2°C has a volume of 536 ml at 637 torr, what will its volume be if the pressure is increased to 712 torr?

Answer 1

Answer:

V = 479.6 mL

Explanation:

assuming ideal gas:

• PV = RTn

∴ T = 25.2°C ≅ 298.2 K.........remains constant

∴ P1 = 637 torr = 0.8382 atm

∴ V1 = 536 mL = 0.536 L

∴ R = 0.082 atm.L/K.mol

⇒ n = (P1V1)/(RT) = ((0.8382 atm)*(0.536 L))/((0.082 atmL/Kmol)*(298.2K))

⇒ n = 0.0184 mol......remains constant

∴ P2 = 712 torr = 0.936842 atm

⇒ V2 = RTn/P2 = [(0.082atmL/Kmol)*(298.2K)*(0.0184mol)]/(0.936842atm)

⇒ V2 = 0.4796 L

Answer 2

The volume will be 0.4796 L, if the pressure is increased to 712 torr.

 Assume the ideal gas situation,  

$\bold {pV = nRT}$

Where,

T -temperature =  298.2 K  

P1 - initial pressure = 637 torr = 0.8382 atm  

V1 - initial volume = 536 mL = 0.536 L  

R - gas constant  = 0.082 atm.L/K.mol

P2- final pressure  = 712 torr = 0.936842 atm

Put values in the formula,  

$\bold {n = (P1V1)/(RT)}\\\\\bold {n = \dfrac {0.8382\ atm \times 0.536 L}{0.082\ atmL/Kmol\times 298.2K}}\\\\\bold {n = 0.0184\ mol}$  

So, the final volume of the gas,  

$\bold {V_2 = \dfrac {nRT}{P2}}\\\\\bold {V_2 = \dfrac {(0.082atmL/Kmol)\times (298.2K)\times (0.0184mol)]}{(0.936842atm)}}\\\\\bold { V2 = 0.4796 L}$

Therefore, the volume will be 0.4796 L if the pressure is increased to 712 torr.

To know more about ideal gas,

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