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One mole of a monoatomic ideal gas is ta...

One mole of a monoatomic ideal gas is taken through the cycle ABCDA as shown in the figure.
`T_A=1000K` and `2p_A=3p_B=6p_C`

`[Ass um e(2/3)^0.4=0.85` and `R=(25)/(3)JK^-1mol^-1]`
The temperature at B is

A

350K

B

1175K

C

850K

D

577K

Text Solution

Verified by Experts

The correct Answer is:
C
In adiabatic process,
`p^(1-gamma)T^gamma=`constant or `Tpropp^((gamma-1)/(gamma))`
`:.` `(T_B)/(T_A)=((p_B)/(p_A))^((gamma-1)/(gamma)`
`:.` `T_B=(1000)(2/3)^((5//3-1)/(5//3))`
`=(1000)(2/3)^0.4=850K`
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Knowledge Check

  • One mole of a monoatomic ideal gas is taken through the cycle ABCDA as shown in the figure. T_A=1000K and 2p_A=3p_B=6p_C [Ass um e(2/3)^0.4=0.85 and R=(25)/(3)JK^-1mol^-1] Work done by the gas in the process ArarrB is

    A
    (a) 5312J
    B
    (b) 1875J
    C
    (c) 6251J
    D
    (d) 8854J

  • One mole of a monoatomic ideal gas is taken through the cycle ABCDA as shown in the figure. T_A=1000K and 2p_A=3p_B=6p_C [Ass um e(2/3)^0.4=0.85 and R=(25)/(3)JK^-1mol^-1] Heat lost by the gas in the process BrarrC is

    A
    (a) 5312J
    B
    (b) 1875J
    C
    (c) 6251J
    D
    (d) 8854J

  • 0.2 moles of an ideal gas is taken round the cycle ABC as shown in the figure. The path B rarr C is an adiabatic process A rarr B is an isochoric process and C rarr A is an isobaric process. The temperature at A and B are T_(A) 300 K and T_(B) = 500 k and pressure at A is 1 atm and volume at A is 4.91. The volume at C is (given : gamma = (C_(P))/(C_(V)) = (5)/(3), R = 8.205 xx 10^(-2) L "atm mol"^(-1) K^(-1), ((3)/(2))^(2//5) = 0.81 )

    A
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    D
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