Question

A propane tank of 1-m3 is at 150 kPa, 300K and connected by a valve to another tank of 0.5 m3 with propane at 250 kPa, 380 K. The valve is opened, and the two tanks come to a uniform state at 320 K. What is the final pressure?

Answer 1

Answer:

Final pressure of the gas 176.55 kPa.

Explanation:

Consider propane as an ideal gas as P << critical pressure.

The ideal gas equation is:

P V = m Rₐ T

Here P is pressure, V is volume, m is mass, T is temperature and Rₐ is individual gas constant.

Rₐ = $\frac{R}{M}$

Here R is universal gas constant and M is mass of propane gas.

Substitute 8.314 J mol⁻¹ K⁻¹ for R and 44.1 x 10⁻³ kg mol⁻¹ for M in the above equation.

Rₐ = $\frac{8.314}{44.1\times10^{-3} }$ = 188.52 J kg⁻¹ K⁻¹

According to the question,

m₁ = $\frac{P_{1}V_{1} }{R_{a} T_{1} }$

Substitute 1 m³ for V₁, 150 x 10³ Pa for P₁, 300 K for T₁ and 188.52 J kg⁻¹ K⁻¹ for Rₐ in the above equation.

m₁ = $\frac{1\times150\times10^{3} }{300\times188.52}$

m₁ = 2.65 kg

According to the question,

m₂ = $\frac{P_{2}V_{2} }{R_{a} T_{2} }$

Substitute 0.5 m³ for V₂, 250 x 10³ Pa for P₂, 380 K for T₂ and 188.52 J kg⁻¹ K⁻¹ for Rₐ in the above equation.

m₂ = $\frac{0.5\times250\times10^{3} }{380\times188.52}$

m₂ = 1.74 kg

After opening the valve, mass of the gas is:

m = m₁ + m₂ = ( 2.65 + 1.74 ) kg = 4.39 kg

Volume of the gas is :

V = V₁ + V₂ = ( 1 + 0.5 ) m³ = 1.5 m³

The temperature of the gas becomes 320 K. So, the final pressure of the gas is given by the relation :

P = $\frac{mR_{a} T }{V} }$

P = $\frac{4.39\times188.52\times320 }{1.5} }$

P = 176.55 x 10³ Pa = 176.55 kPa