Question

7. A cylinder contains 0.45 m3 of a gas at 1 × 105 N/m2 and 80°C. The gas

is compressed to a volume of 0.13 m3, the final pressure being 5 × 105 N/m2. Determine :

(i) The mass of gas ;

(ii) The value of index ‘n’ for compression ;

(iii) The increase in internal energy of the gas ;

(iv) The heat received or rejected by the gas during compression.

Take γ = 1.4, R = 294.2 J/kg°C

Answer 1

$V_1=0.45 \;m^3 \\ P_1 = 1 \times 10^5 \;N/m^2 \\ T_1=80 \;°C \\ = 80+273=353 \;K \\ V_2 = 0.13 \;m^3 \\ P_2 = 5 \times 10^5 \;N/m^2$

(i) The mass of gas

$P_1V_1 = mRT_1 \\ m=\frac{1 \times 10^5 \times 0.45 \times 10^{-3}}{0.294 \times 353} \\ m=0.43360 \;kg$

(ii) The value of index “n” for compression

$(\frac{P_2}{P_1})^{\frac{n-1}{n}} = (\frac{V_1}{V_2})^{n-1} \\ (\frac{P_2}{P_1})^{\frac{1}{n}}= \frac{V_1}{V_2} \\ (\frac{5 \times 10^5}{1 \times 10^5})^{\frac{1}{n}}= \frac{0.45}{0.13} \\ \frac{1}{n}ln(5) = ln(3.46153) \\ n= \frac{ln(5)}{ln(3.46153)} \\ n= \frac{1.6094}{1.2417106} \\ n=1.2961 ≈1.3 \\ n=1.3$

(iii) The increase in internal energy of the gas

$ΔU =mC_v(T_2-T_1) \\ C_v= \frac{R}{k-1} \\ ΔU = \frac{mR(T_2-T_1)}{k-1} \\ P_2V_2 =mRT_2 \\ P_1V_1 =mRT_1 \\ ΔU = \frac{P_2V_2-P_1V_1}{k-1} \\ = \frac{(5 \times 10^5 \times 0.13) -(1 \times 10^5 \times 0.45)}{1.4-1} \\ = 50000\;J$

(iv) The heat received or rejected by the gas during compression.

$Q_{1-2} = ΔU+W_{1-2} \\ W_{1-2}=\frac{P_1V_1 -P_2V_2}{n-1}=-67567.567 \;J \\ Q_{1-2}= 50000-67567.567=-17567.567 \;J$

Heat was rejected.