Two moles of Helium gas (γ=5 / 3) are initially at temperature 27° C and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value. What is the work done by the gas? (Gas constant R =8.3 T / mole K )

Question

Two moles of Helium gas (&gamma;=5 / 3) are initially at temperature 27° C and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value. What is the work done by the gas? (Gas constant R =8.3 T / mole K ) (A) 13450 J (B) 14450 J (C) 16450 J (D) 12450 J

Tardigrade Admin · Accepted Answer

The work done by gas in isobaric process

A B

= 2.49 \times 1 0^{5} \times (40 - 20) \times 1 0^{- 3} = 4980 J

The work done by gas during adiabatic process

BC

W_{2} = \frac{n R}{1 - γ} [T_{2} - T_{1}] = \frac{2 \times 8.3}{1 - ( 5/3 )} [300 - 600]

= 7470 J .

∴

Net work done by gas

W = W_{1} + W_{2}

= 4980 + 7470 = 12450 J

$13450 J$

$14450 J$

$16450 J$

$12450 J$

13450J

14450J

16450J

12450J

The work done by gas in isobaric process AB =2.49×105×(40−20)×10−3=4980J The work done by gas during adiabatic process BC W2​=1−γnR​[T2​−T1​]=1−(5/3)2×8.3​[300−600] =7470J. ∴ Net work done by gas W=W1​+W2​ =4980+7470=12450J

$13450 J$

$14450 J$

$16450 J$

$12450 J$