Chemistry
Bowline347
37

The gas in a cylinder has a volume of 8 liters at a pressure of 101 kPa. The pressure of the gas is increased to 213 kPa. Assuming the temperature remains constant, what would the new volume be? Please show your work, thanks!

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(1) Answers
Alex921

We'll use the Ideal Gas Law. Before: [latex]p_1V_1=n_1RT_1\\\\ (101\times10^{3})\cdot8=n_1RT_1\\\\ n_1RT_1=808\times10^3~~~(i)[/latex] After: [latex]p_2V_2=n_2RT_2\\\\ (213\times10^3)\cdot V_2=n_2RT_2~~~(ii)[/latex] Since the quantity of gas wasn't changed, [latex]n_1=n_2=n[/latex]. Furthermore, the temperature remains constant, so [latex]T_1=T_2=T[/latex]. Then, the expressions that were find are: [latex]nRT=808\times10^3~~~(i)\\\\ nRT=(213\times10^3)\cdot V_2~~~(ii)[/latex] Dividing them: [latex]\dfrac{nRT}{nRT}=\dfrac{808\times10^3}{(213\times10^3)\cdot V_2}\\\\ 1=\dfrac{808}{213\cdot V_2}\\\\ 213V_2=808\\\\ V_2=\dfrac{808}{213}\\\\ \boxed{V_2\approx 3.79~L}[/latex]

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