A) 20 cal
B) 40 cal
C) 60 cal
D) 80 cal
Correct Answer: A
Solution :
| [a] At constant pressure |
| \[{{(\Delta Q)}_{p}}=\mu {{C}_{p}}\Delta T=1\times {{C}_{p}}\times (30-20)=40\] |
| \[\Rightarrow \] \[{{C}_{p}}=4\frac{\text{calorie}}{\text{moke}\,\text{kelvvin}}\] |
| \[\therefore \] \[{{C}_{v}}={{C}_{p}}-R=4-2=2\,\frac{\text{calorie}}{\text{mole }\!\!\times\!\!\text{ kelvin}}\] |
| Now, \[{{(\Delta Q)}_{v}}=\mu {{C}_{v}}\,\Delta T\] |
| \[=1\times 2\times (30-20)\] |
| \[=20\,\,\text{cal}\] |
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