Home
Class 11
CHEMISTRY
One mole of a gas changed from its initi...

One mole of a gas changed from its initial state (15L,2 atm) to final state (4L,10 atm) reversibly. If this change can be represented by a straight line in P - V curve maximum temperature (approximate), the gas attained is `x xx10^(2) K.` Then find the value of x.

Text Solution

Verified by Experts

The correct Answer is:
7
Equation of line
`P-2=(10-2)/(4-15)(V-15)`
`P-2=(8)/(11)(V-15)`
`P=2-(8V)/(11)+(15xx8)/(11)`
`P=((142)/(11)-(8V)/(11))`

`f(T)=(1)/(nR)((142V)/(11)-(8V^(2))/(11))`
`(d(F(T)))/(dV)=(1)/(nR)((142)/(11)-(8V^(2))/(11))=0`
`V=(142)/(11xx16)=8.875`
`P=(142)/(11)-(8)/(11)xx(8.875)/(16)=(71)/(11)`
`T_("Max")=(PV)/(nR)=(71)/(11)xx(142)/(16xx0.0821)=700`
`=7xx10^(2)K`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • GASEOUS STATE

    NARENDRA AWASTHI|Exercise Level 3 - Passage|1 Videos

  • ELECTROCHEMISTRY

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|1 Videos

  • IONIC EEQUILIBRIUM

    NARENDRA AWASTHI|Exercise Assertin-Reason Type Questions|1 Videos

Similar Questions

Explore conceptually related problems

A gas takes part in two thermal processes in which it is heated from the same initial state to the same final temperature. The P-V diagram for these two processes are indicated by straight lines 1-3 and 1-2 in figure. Find out in which process the amount of heat supplied to the gas is larger.

For a reaction the graph drawn between half life period and reciprocal of initial concentration(1/a) gives a straight line with positive slope. The magnitude of the slope is 500 and iniital concentration of the reactant is 2M. If the initial rate of the reaction is represented as x xx10^(-3) units, then the value of x is _____________

Knowledge Check

  • One mole of an ideal monoatomic gas at 27^(@)C is subjected to a reversible isoentropic compression until final temperature reaches 327^(@)C . If the initial pressure was 1.0 atm then find the value of ln P_(2) :

    A
    1.75
    B
    0.176
    C
    1.0395
    D
    `2.0`

  • 1 L of oxygen at a pressure of 1 atm and 2 L of nitrogen at a pressure of 0.5 atm, are introduced into a vessel of volume 1 L. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is

    A
    1.5
    B
    2.5
    C
    2
    D
    4

  • 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) ]

    A
    0
    B
    R ln (24.6)
    C
    R ln (2490)
    D
    `(3)/(2)` R ln (24.6)

    • Similar Questions

    Explore conceptually related problems

    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) ]

    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:

    One mole of a gas expands such that its volume 'V' changes with absolute temperature 'T' in accordance with the relation V = KT^2 where 'K' is a constant. If the temperature of the gas changes by 60^@C , then work done by the gas (R is universal gas constant)

    One mole of a gas expands such that its volume 'v' changes with absolute temperature 'T' in accordance with the ralation V = HT^(2) where 'K' is a constant . If the temperature of the gas changes by 60^(@) C , then work done by the gas is (R is universal gas constant ) .

    One mole of nitrogen gas being initially at a temperature of T_0 =300 K is adiabatically compressed to increase its pressure 10 times. The final gas temperature after compression is (Assume nitrogen gas molecules as rigid diatomic and 100^(1/7) =1.9)

    Home
    Profile