One mole of an ideal gas at 300 K is expanded Isothermally from an initial volume of 1 L to 10 L. The Δ E for this process is (R=2. cal mol .-1 K-1)

Question

One mole of an ideal gas at 300 K is expanded Isothermally from an initial volume of 1 L to 10 L. The Δ E for this process is (R=2. cal mol .-1 K-1) (A) 163.7 cal (B) zero (C) 1381.1 cal (D) 9 L atm

Tardigrade Admin · Accepted Answer

For a finite change Δ E=Cv Δ T. For an isothermal process Δ T=0. Hence, Δ E=0.

Q. One mole of an ideal gas at $300 K$ is expanded Isothermally from an initial volume of $1 L$ to $10 L$ . The $Δ E$ for this process is $(R = 2$ cal mol $^{- 1} K^{- 1})$

$163.7 c a l$

zero

$1381.1 c a l$

$9 L a t m$

For a finite change $Δ E = C_{v} Δ T$ . For an isothermal process $Δ T = 0$ . Hence, $Δ E = 0$ .

Q. One mole of an ideal gas at 300K is expanded Isothermally from an initial volume of 1L to 10L. The ΔE for this process is (R=2 cal mol −1K−1)

163.7cal

zero

1381.1cal

9Latm

For a finite change ΔE=Cv​ΔT. For an isothermal process ΔT=0. Hence, ΔE=0.

Q. One mole of an ideal gas at $300 K$ is expanded Isothermally from an initial volume of $1 L$ to $10 L$ . The $Δ E$ for this process is $(R = 2$ cal mol $^{- 1} K^{- 1})$

$163.7 c a l$

$1381.1 c a l$

$9 L a t m$

For a finite change $Δ E = C_{v} Δ T$ . For an isothermal process $Δ T = 0$ . Hence, $Δ E = 0$ .