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When 1mol of a monoatomic ideal gas at T...

When `1mol` of a monoatomic ideal gas at `TK` undergoes adiabatic change under a constant external pressure of `1atm`, changes volume from `1 L to 2L`. The final temperature (in K) would be

A

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

B

`T+2/(3xx0.0821)`

C

T

D

`T- 2/(3xx0.0821)`

Text Solution

Verified by Experts

The correct Answer is:
D
Work done by the gas `W = - P Delta V`.
For irreversible adiabatic expansion.
`W = C_V (T_2 - T_1) , C_V = 3//2 R` for monoatomic gas
Thus ` - P Delta V = C_V (T_2 - T_1)`
` - 1 xx (2 - 1) = (3//2) R (T_2 - T_1)`
`T_2 - T_1 (-2)/(3R) rArr T_2 = T_1 - 2/(3 xx 0.0821)`
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