Two moles of helium (γ = 5/3) expand adiabatically so that its temperature falls from 100°C to 50°C. Then the work done by the gas will be

Question

Two moles of helium (&gamma; = 5/3) expand adiabatically so that its temperature falls from 100°C to 50°C. Then the work done by the gas will be (A) 50 R (B) 60 R (C) 100 R (D) 150 R

Tardigrade Admin · Accepted Answer

When a gas is expand adiabatically, work done by the gas is

W = \frac{n R}{( γ - 1 )} (T_{1} - T_{2})

Here,

n = 2, γ = \frac{5}{3}

T_{1} = 10 0^{\circ} C

and

T_{2} = 5 0^{\circ} C

∴ W = \frac{2 R}{( \frac{5}{3} - 1 )} (100 - 50)

= \frac{2 R}{2/3} (50)

150 R

Q. Two moles of helium $(γ = 5/3)$ expand adiabatically so that its temperature falls from 100°C to 50°C. Then the work done by the gas will be

$50 R$

$60 R$

$100 R$

$150 R$

When a gas is expand adiabatically, work done by the gas is
$W = \frac{n R}{( γ - 1 )} (T_{1} - T_{2})$
Here, $n = 2, γ = \frac{5}{3}$
$T_{1} = 10 0^{\circ} C$ and $T_{2} = 5 0^{\circ} C$
$∴ W = \frac{2 R}{( \frac{5}{3} - 1 )} (100 - 50)$
$= \frac{2 R}{2/3} (50)$
$150 R$

Q. Two moles of helium (γ=5/3) expand adiabatically so that its temperature falls from 100°C to 50°C. Then the work done by the gas will be

50R

60R

100R

150R