Answer to Question #182213 in Molecular Physics | Thermodynamics for Paras

Question

A cylinder contains 0.5 m3 of a gas at 1 × 105 N/m2 and Z temperature in °C (For Z value refer Table. 1). The gas is compressed to a volume of 0.125 m3 , the final pressure being 6 × 105 N/m2 . Determine: (i) The mass of gas. (ii) The value of index ‘n’ for compression. (iii) The increase in internal energy of gas. (iv) The heat received or rejected by the gas during compression

Answer 1

$V_1 = 0.5m³\\ P_1 = 1× 10⁵\ N/m²\\ T_1 = 2°C +273 = 275K$

$V_2 = 0.125m³\\ P_2 = 6× 10⁵N/m²$

(i)

$m =\dfrac{P_1V_1}{RT_1} = \dfrac{1×10⁵×0.5}{294.2×275}=$

$0.618kg$

(ii)

$(\dfrac{V_1}{V_2})^n = (\dfrac{P_2}{P_1})$

$(\dfrac{0.5}{0.125})^n = (\dfrac{6×10⁵}{1×10⁵})$

$4^n = 6\\ n = \log_46 = 1.3$

(iii)

In a polyprotic process,

$(\dfrac{T_2}{T_1}) = (\dfrac{V_1}{V_2})^{n-1} = 4^{1.3-1} = 4^{0.3} =$

$1.516$

$(\dfrac{T_2}{T_1}) =1.516$

$T_2 = 1.516 ×275 = 413.8K$

$∆U = mc_v (T_2-T_1)$

$∆U = 0.618 ×\dfrac{294.2}{(1.4-1)}×138.8$

$∆U =63000J= 63kJ$

(iv)

$W = \dfrac{mR(T_1-T_2)}{n- 1}$

$W = \dfrac{0.618× 294.2× (-138.8)}{1.3-1}$

$W = -84120J = -84.12kJ$

$Q = ∆U + W\\ Q = 63 +(- 84.12) =-21.2kJ$

-21.2kJ heat is rejected by the gas during compression.