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2 mol of an ideal monatomic gas is carri...

2 mol of an ideal monatomic gas is carried from a state `(P_(0), V_(0))` to a state `(2P_(0), 2V_(0))` along a straight line path in a P - V diagram. The amount of heat absorbed by the gas in the process is given by

A

`3P_(0)V_(0)`

B

`9/2P_(0)V_(0)`

C

`6P_(0)V_(0)`

D

`3/2P_(0)V_(0)`

Text Solution

Verified by Experts

The correct Answer is:
C
According to the graph, if temperatures of state A and state B of the gas are `T_(1) and T_(2)`, then
`P_(0)V_(0) = nRT_(1) and 2P_(0) xx 2V_(0) = nRT_(2)`
`:. T_(2) - T_(1) = (4 P_(0)V_(0))/(nR) - (P_(0)V_(0))/(nR) = (3P_(0)V_(0))/(nR)`
From state A to B at intermediate state, the pressure is changed from `P_(0)` to `2P_(0)` by changing the temperature from `T_(1)` to `T_(2)` keeping volume constant.

Then increase in internal energy of the gas,
`Delta U` = the amount of heat absorbed by n mol gas at constant volume
`= n xx C_(v) xx (T_(2) - T_(1))`
`= n xx (3R)/2 xx (3P_(0)V_(0))/(nR)` [ for monatomic gas, `C_(v) = (3R)/2]`
`= (9P_(0)V_(0))/(2)`
Work done by the gas absorbing the remaining heat,
`Delta W = "area of" ABCD`
` = (2 V_(0) + V_(0)) (2P_(0) - P_(0))/(2) = (3P_(0)V_(0))/(2)`
`:. Delta Q = Delta U + Delta W = 9/2 P_(0)V_(0) + 3/2 P_(0)V_(0) = 6P_(0)V_(0)`.
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