A) 409.5 J
B) 819 J
C) 1638 J
D) 3276 J
Correct Answer: C
Solution :
At constant volume Pressure \[P\propto T\] to double the pressure, temperature should be doubled. Initial temperature = 273K (at S.T.P.) Final temperature \[=2\times 273=546\text{ }K\] Rise in temperature \[\Delta t=546-273=273K\] Mass of \[\frac{1}{2}\] mole \[=\frac{1}{2}\times 4=2g\] Heat \[Q=ms\Delta t\] \[=2\times 3\times 273=1638J\]You need to login to perform this action.
You will be redirected in
3 sec