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1mol of an ideal gas undergoes reversibl...

`1mol` of an ideal gas undergoes reversible isothermal expansion form an initial volume `V_(1)` to a final volume `10V_(1)` and does `10kJ` of work. The initial pressure was `1xx 10^(7) Pa`.
a. Calculate `V_(2)`.
b. If there were `2mol` of gas, what must its temperature have been?

A

0.003, 275 K

B

0.00043, 261.13 K

C

0.0005, 300 K

D

0.00049, 353 K

Text Solution

AI Generated Solution

To solve the given problem step by step, we will break it down into two parts as per the question. ### Part (a): Calculate \( V_2 \) 1. **Understanding the Process**: We have 1 mole of an ideal gas undergoing a reversible isothermal expansion from an initial volume \( V_1 \) to a final volume \( V_2 = 10V_1 \). The initial pressure \( P_1 \) is given as \( 1 \times 10^7 \) Pa. 2. **Using the Ideal Gas Law**: The ideal gas law is given by: \[ ...
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Knowledge Check

  • The work done, W, during an isothermal process in which the gas expands from an intial volume V_(1) , to a final volume V_(2) is given by (R : gas constant, T : temperature )

    A
    `R (V_(2)-V_(1)) log_(e _ ((T_(1))/(T_(2))`
    B
    `R (T_(2)-T_(1)) log_(e ) ((V_(1))/(V_(2)))`
    C
    `RT log_(e ) [(V_(2))/(V_(1))]`
    D
    `2RT log_(e ) [(V_(1))/(V_(2))]`

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