One mole of an ideal monoatomic gas at t...

One mole of an ideal monoatomic gas at temperature T and volume 1L expands to 2L against a constant external pressure of one atm under adiabatic conditions, then final temperature of gas will be:

`(T)/(2^(2//3))`

`T + (2)/(3 xx 0.0821)`

`T - (2)/(3 xx 0.0821)`

Text Solution

Verified by Experts

The correct Answer is:

Constant `P rArr` irreversibe
`Delta E= - P_("ext") Delta V`
`nC_(V) Delta T = -P_("ext") (V_(2) - V_(1))`
` 1 xx (3)/(2) R xx (T_(2) - T_(1)) = -1(2-1) = -1`
`T_(2) = T- (2)/(3R)`

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One molar of an ideal monoatomic gas at temperature T_0 expands slowly according to the law P/V = constant. If the final temperature is 2T_0 heat supplied to the gas is

`2RT_0`

`3/2 RT_0`

`RT_0`

`1/2 RT_0`

One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1l to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300K then total entropy change of system in the above process is: [R =0.082L atm mol^(-1)K^(-1)= 8.3J mol^(-1) K^(-1) ]

R ln (24.6)

R ln (2490)

`(3)/(2)` R ln (24.6)

One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300K then total entropy change of system in the above process is [R = 0.082L atm mol^(-1) K^(-1) = 8.3J mol^(-1) K^(-1) ]