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Mohan Aatif Wable
Asked: 2022-11-11T14:08:33+05:30 In:

For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using expression ` = -nRT ln.(V_(f))/(V_(i))`. A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversible to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option
A. Work done at 600 K is 20 times the work done at 300 K
B. work done at 300K is twice the work done at 600 K
C. work done at 600 K is twice the work done at 300 K
D. `Delta U = 0` in both cases

For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using expression ` = -nRT ln.(V_(f))/(V_(i))`. A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversible to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option
A. Work done at 600 K is 20 times the work done at 300 K
B. work done at 300K is twice the work done at 600 K
C. work done at 600 K is twice the work done at 300 K
D. `Delta U = 0` in both cases
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  1. Correct Answer – C::D
    `w = -nRT ln.(V_(f))/(V_(i))` Here, n, R and `ln.(V_(f))/(V_(i))` are same in both the cases therefore
    `(w_(2))/(w_(1)) = (T_(2))/(T_(1))` or `(w(600K))/(w(300K)) = (600K)/(300K) = 2`

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Mini Lalit Johal
Asked: 2022-11-08T19:51:59+05:30 In:

For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression `w= – nRT ln . (V_(f))/( V_(i)0` . A sample containig 1.0 molof an ideal gas is expanded isothermally and reversibly to ten times of its original volume in two separate experiments . The expansions is carried out at 300 K and at600 K respectively.
Choose the correct option.
A. Work done at 600 K is 20 times at the work done at 300K
B. Work done at 300 K is twice the work done at 600 K
C. Work done at 600 K is twice the work done at 300 K
D. `DeltaU=0`in both cases

For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression `w= – nRT ln . (V_(f))/( V_(i)0` . A sample containig 1.0 molof an ideal gas is expanded isothermally and reversibly to ten times of its original volume in two separate experiments . The expansions is carried out at 300 K and at600 K respectively.
Choose the correct option.
A. Work done at 600 K is 20 times at the work done at 300K
B. Work done at 300 K is twice the work done at 600 K
C. Work done at 600 K is twice the work done at 300 K
D. `DeltaU=0`in both cases
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  1. Correct Answer – c,d
    `w= -nRT ln. (V_(1))/( V_(2)). `The factors n , R and `ln. (V_(f))/( V_(i))` are same in both the cases. Hence, `(w_(2))/(w_(1)) = (T_(2))/(T_(1)), i.e., (.^(w)600K)/(.^(w)300K)= ( 600K)/( 300K) =2 , i.e., ` work done at 600K is twice the work done at 300K.
    As each case involves isothermal expansion of an ideal gas, there is no change in internal energy, i.e.,
    `DeltaU =0`

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Darpan Dev Kakar
Asked: 2022-11-06T20:19:44+05:30 In:

For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using expression ` = -nRT ln.(V_(f))/(V_(i))`. A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversible to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option
A. work done at 600 K is 20 times the work done at 300 K
B. work done at 300 K is twice the work done at 600 K
C. work done at 600 K is twice the work done at 300 K
D. `Delta U = 0` in both cases.

For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using expression ` = -nRT ln.(V_(f))/(V_(i))`. A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversible to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option
A. work done at 600 K is 20 times the work done at 300 K
B. work done at 300 K is twice the work done at 600 K
C. work done at 600 K is twice the work done at 300 K
D. `Delta U = 0` in both cases.
Parabola
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  1. Correct Answer – C::D
    Given that, the work of reversible expansion under isothernal condition can be calculated by using the expression
    `W = – nRT ln. (V_(1))/(V_(1))`
    `V_(1) = 10 V_(i)`
    `T_(2) = 600 K`
    `T_(1) = 300 K`
    Putting these values in above expression
    `W_(600 K) = 1 xx R xx 600 ln. (10)/(1)`
    `W_(300 K) = 1 xx R xx 300 K ln. (10)/(1)`
    Ration `= (W_(600 K))/(W_(300 K)) = (1 xx R xx 600 K ln.(10)/(1))/(1 xx R xx 300 K ln. (10)/(1)) = (600)/(300) = 2`
    For isothermal expansion of ideal gases, `Delta U = 0`. Since, temperature is contant this means there is no change in internal energy. Therefore, `Delta U = 0`

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