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Five moles of hydrogen initially at STP ...

Five moles of hydrogen initially at STP is compressed adiabatically so that its temperature becomes 673 K. The increase in internal energy of the gas , in kilo joule is (R= 8.3 J`//` mol-K, `gamma`= 1.4 for diatomic gas)

A

80.5 kJ

B

21.55 kJ

C

41.50 kJ

D

65.55 kJ

Text Solution

Verified by Experts

The correct Answer is:
C
Given, temperature, `T_(2) = 673 K `
initial temperature `T_(1) = 273 ` K
rate of heat flow, R = 8.3 J `mol^(-1) K^(-1)`
`gamma` = 1.4 for diatomic gas, moles of hydrogen gas,
Now, change in internal energy due to adiabatic process is
`therefore " " Delta U = nC_(v) Delta T `
or `" " Delta U = 5 xx (R )/(gamma -1) (T_(2) - T_(1)) `
`( because "Specific heat", C_(v) = (R )/(gamma - 1) )`
or `" " = 5 xx (8.3)/(1.4 - 1) (673 - 273) `
or `" " = 5 xx (8.3)/(0.4) xx 400 `
or ` Delta U = 41.5 ` k,J
So the increase in internal energy of the gas is `Delta U` = 41.54 KJ.
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