Question

A sample of helium measuring 6L was kept at a pressure of 1.5 atm 15° C. If the pressure is 3atm and 10°C, what would be its new volume?

Answer 1

Given:

• Initial pressure(p1) = 1.5 atm.

• Final pressure(p2) = 3 atm.

• Initial volume(v1) = 6 L.

• Initial Temperature(T1)= 15° C.

• Final temperature(T2) = 10° C.

To find:

The final volume(v2) of the gas.

Solution:

T1 = 15° C

   = (15+273.15) k

   = 288.15 K

Hence, we get, T1 = 288.15 K

Similarly,

T2 = 10° C

   = ( 10 + 273.15) K

  = 283.15 K

Hence, we get, T2= 283.15 K

From the ideal gas law equation we get that,

$\dfrac{p1v1}{T1} =\dfrac{p2v2}{T2}$

⇒ (1.5×6)/288.15 = (3×v2)/283.15

⇒  (3×v2)/283.15 =( 1.5×6)/288.15

⇒ v2 = (1.5×6×283.15) / (288.15×3)

⇒ v2 = (1.5×6×28315) / (28815 × 3)

⇒ v2 = 2.94794377928

⇒ v2 = 3 L approx.

∴  The new volume would be 3 L.

Answer:

              The new volume would be 3 L.

Answer 2

Answer:

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