Question

One litre of Helium under a pressure of 1
atm and at a temperature of 0°C
is
heated until the pressure becomes 1/3
and volume are doubled. The final
temperature attained by the gas is​

Answer 1

Answer:

from ideal gas equation, PV = nRT

So, T=

nR

PV

Since, n is constant, and R is also constant, we can say that T is directly proportional to the product of P and V

TαPV

Since both volume and pressure are doubled, absolute temperature of the gas will become 4 times

Initial temperature = 27

∘

C = 300K

So, final temperature = 4 x 300K = 1200 K = 927

∘

C

Answer 2

Concept:

The Ideal gas equation mathematically expressed as PV = kT should be followed.

Given:

Initial Pressure, P1 = 1 atm

Initial Temperature, T1 = 0⁰C = 273 K

Volume, V1 = 1 litre

Final pressure, P2 = 1/3 atm

Volume = 2V1 = 2(1) = 2 litre

Find:

We need to determine the final temperature, T2 attained by the gas.

Solution:

For an ideal gas, PV = kT where  P = Pressure, V = Volume, T = temperature and k = constant

Therefore, Initially, the equation should be P1V1 = kT1

P1V1/T1 = k ---- equation 1

and final equation should be P2V2 = kT2

P2V2/ T2 = k ---- equation 2

Since R.H.S of both equations is the same therefore L.H.S should also be the same.

P1V1/T1 = P2V2/ T2

We need to find T2 therefore equation can be expressed as-

T2 = P2V2T1 / P1V1

T2 = 1/3×2×273 / 1×1

T2 = 182 K

In celcius it becomes, 182 - 273 = -91⁰C

Thus, the final temperature attained by the gas is -91⁰C.

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