One mole of an ideal gas requires 207 J heat to raise it’s temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume the heat required to raise the temperature by 10 K for the same gas, will be : (given R=8.3 J/mol/K)

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29 J

124 J

198.7 J

215.3 J

answer is B.

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Detailed Solution

$\begin{array}{rcl} a m o u n t o f h e a t r e q u i r e d t o r a i s e t h e t e m p e r a t u r e o f g a s a t c o n s \tan t p r e s s u r e i s Q_{p} & = & n C_{p} ∆ T \end{array}$

$\begin{array}{rcl} g i v e n Q_{p} & = & 207 J o u l e; n u m b e r o f m o l e n = 1; c h a n g e i n t e m p e r a t u r e ∆ T = 10 K \end{array}$

$\begin{array}{rcl} s u b s t i t u t e t h e g i v e n v a l u e s i n a b o v e e q u a t i o n o f Q_{p} \\ 207=1 \times C_{p} \times 10 . . . . . . . . . (1) \\ a m o u n t o f h e a t r e q u i r e d t o r a i s e t h e t e m p e r a t u r e o f g a s a t c o n s \tan t v o l u m e i s Q_{v} & = & n C_{v} ∆ T \\ g i v e n n & = & 1; ∆ T = 10 K, s u b s t i t u t e i n Q_{v} \\ Q_{v} =1 \times C_{v} \times 10 . . . . . . . . (2) \\ s u b t r a c t, e q n (1) - e q n (2), \\ ({207-Q}_{v}) =10 (C_{p} - C_{v}) \\ \Rightarrow & {207-Q}_{v} =10 \times 8.3=83 \\ \Rightarrow & Q_{v} =207-83 \\ \Rightarrow & Q_{v} =124 J \end{array}$