In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas ( $\gamma=5/3$ ) decreases from $60$ K to $50$ K. The work done by the gas in the process is :
(Take the universal gas constant as $R=8.3$ J mol $^{-1}$ K $^{-1}$ )

Question

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas (

\gamma=5/3

) decreases from

60

K to

50

K. The work done by the gas in the process is :
(Take the universal gas constant as

R=8.3

J mol

^{-1}

K

^{-1}

)

Quantrex Academy · Accepted Answer

Correct answer: D. 124.5 J

The work done by an ideal gas in an adiabatic process is given by the formula:

W = \dfrac{nR(T_1 - T_2)}{\gamma - 1}

Given values are:

n = 1

mole

R = 8.3

J mol

^{-1}

K

^{-1}

T_1 = 60

K

T_2 = 50

K

\gamma = \dfrac{5}{3}

Substituting these values into the formula:

W = \dfrac{1 \times 8.3 \times (60 - 50)}{\dfrac{5}{3} - 1}

W = \dfrac{8.3 \times 10}{\dfrac{2}{3}}

W = \dfrac{83 \times 3}{2}

W = \dfrac{249}{2} = 124.5

J

Answer:

124.5

J

In an adiabatic expansion, the temperature of one mole of an ideal...