Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
The work done in compressing the gas by factor 11 is 12,100 J.
The given parameters;
Let the initial volume of the gas = v₁
The work done in compressing the gas to half its initial volume and by factor of 11 is calculated as follows;
[tex]W = P\Delta v\\\\\frac{W_1}{\Delta v_1} = \frac{W_2}{\Delta v_2} \\\\W_2 = \frac{W_1 \times 11 v_1}{0.5v_1} \\\\W_2 = \frac{550 \times 11}{0.5} \\\\W_2 = 12,100 \ J[/tex]
Thus, the work done in compressing the gas by factor 11 is 12,100 J.
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