An ideal gas has a specific heat at constant pressure C_P= (5/2)R. The gas is kept in a closed vessel of volume 8.3 x $10^{- 3}$ $m^{3}$ at a temperature of 300 K and a pressure of 1.6 x $10^{6}$ N/ $m^{2}$ . An amount of 2.49 x $10^{4}$ J of heat energy is supplied to the gas. Calculate the final pressure

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$3 \times 10^{6} N / m^{2}$

$3.2 \times 10^{6} N / m^{2}$

$3.6 \times 10^{6} N / m^{2}$

$n o n e o f t h e s e$

answer is C.

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Detailed Solution

$d i f f e r e n c e o f s p e c i f i c h e a t s i s C_{P} - C_{V} = R C_{V} = C_{P} - R C_{V} = (5 / \underline{2) R} - R C_{V} = (3 / 2) R \begin{array}{l} g i v e n p r e s s u r e P = 1.6 x 10^{6} p a s c a l; v o l u m e V = 8.3 x 10^{- 3} m^{3}; t e m p e r a t u r e T = 300 K; u n i v e r s a l g a s c o n s \tan t = 8.3 J / m o l - K \\ n u m b e o f m o l e s = μ = \frac{P V}{R T} = \frac{1.6 x 10^{6} \times 8.3 x 10^{- 3}}{8.3 \times 300} \\ μ = \frac{16}{3} \\ a m o u n t o f h e a t g i v e n i s, \\ Δ U = (Δ Q)_{V} = μ C_{V} d T (∵ W o r k d o n e, ∆ W = 0 f o r c l o s e d r i g i d c o n t a i n e r a s v o l u m e i s c o n s \tan t) \\ 2.49 x 10^{4} = \frac{16}{3} \times \frac{3}{2} \times 8.3 d T \\ \Rightarrow d T = T_{2} - T_{1} = 375 \\ T_{2} = T_{1} + 375 \\ \Rightarrow T_{2} = 300 + 375 = 675 \\ w e k n o w P α T \\ \frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}} \\ \frac{1.6 x 10^{6}}{P_{2}} = \frac{300}{675} \\ \Rightarrow P_{2} = 3.6 \times 10^{6} N / m^{2} \end{array}$