7 mole of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is (Given R =8.3 JK -1 mol -1 )

Question

7 mole of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is (Given R =8.3 JK -1 mol -1 ) (A) 5810 J (B) 3486 J (C) 11620 J (D) 6972 J

Tardigrade Admin · Accepted Answer

For a quasi-static process the change in internal energy of an ideal gas is

Δ U = n C_{V} Δ T

= n \times \frac{3 R}{2} \times Δ T

[molar heat capacity at constant volume for mono atomic gas

= \frac{3 R}{2}

]

Δ U = 7 \times \frac{3}{2} \times 8.3 \times 40 = 3486 J

Q. $7$ mole of certain monoatomic ideal gas undergoes a temperature increase of $40 K$ at constant pressure. The increase in the internal energy of the gas in this process is (Given $R = 8.3 J K^{- 1} m o l^{- 1}$ )

$5810 J$

$3486 J$

$11620 J$

$6972 J$

For a quasi-static process the change in internal energy of an ideal gas is
$Δ U = n C_{V} Δ T$
$= n \times \frac{3 R}{2} \times Δ T$
[molar heat capacity at constant volume for mono atomic gas $= \frac{3 R}{2}$ ]
$Δ U = 7 \times \frac{3}{2} \times 8.3 \times 40 = 3486 J$

Q. 7 mole of certain monoatomic ideal gas undergoes a temperature increase of 40K at constant pressure. The increase in the internal energy of the gas in this process is (Given R=8.3JK−1mol−1 )

5810J

3486J

11620J

6972J