28g of `N_(2)`gas at 300 K and 20 atm was allowed to expand isothermally against a constant external pressure of 1 atm , q for the gas is `(R=0.082)`.

Question

28g of `N_(2)`gas at 300 K and 20 atm was allowed to expand isothermally against a constant external pressure of 1 atm , q for the gas is `(R=0.082)`.

28g of `N_(2)`gas at 300 K and 20 atm was allowed to expand isothermally against a constant external pressure of 1 atm , q for the gas is `(R=0.082)`.
A. 2495 J
B. 7473 J
C. 2367 J
D. 2570 J

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
Initial volume `(V_1)` for `N_2 (PV_1=n RT)`
`20xxV_1=((28)/(28))xx0.082xx300`
`therefore V_1=1.23L`
Final volume `(V_2)` for `N_2(PV_2=n RT)`
`1xxV_2=((28)/(28))xx0.082xx300`
`rArr V_2=2460L`
Since expansion is made against constant pressure and thus irreversible.
`P=1 atm =1.013xx10^(5)N//m^2`
`Delta V=(V_2-V_1)=(24.60-1.23)L=23.37xx10^(-3)m^(3)`
`therefore W=-PxxDelta V=-1.013xx10^(5) xx23.37xx10^(-3)`
`=-23.37xx10^(2)J`
`therefore q= Delta U-W=0+23.67xx10^(2)=+2367J`
(`because Delta U=0` , in an internal expansion)