Answer to Question #110561 in Chemistry for alyssa

Question

If 10.0 L of oxygen at STP are heated to 512 degrees C what will be the new volume of gas if the pressure is also increased to 202.6kPa?

Answer 1

Solution:

The Ideal Gas Law can be written as: PV = nRT,

where P, V and T are the pressure, volume and temperature; n is the amount of substance; and R is the ideal gas constant.

Assuming 2 set of conditions:

Initial case: P_iV_i = n_iRT_i

Final case: P_fV_f = n_fRT_f

Since the ideal gas constant (R) has the same value in each case, one will get the final equation and the General Gas Equation:

(P_iV_i / n_iT_i) = (P_fV_f / n_fT_f).

In our case, the amount of substance (n) does not change, therefore:

(P_iV_i / T_i) = (P_fV_f / T_f),

or P_iV_iT_f = P_fV_fT_i

P_i = 101.325 kPa (1atm, STP); V_i = 10.0 L; T_i = 273.15 K (0^oC, STP);

P_f = 202.6 kPa; V_f = unknown; T_f = 512^oC = 785.15 K.

P_iV_iT_f = P_fV_fT_i

Then,

(101.325 kPa * 10.0 L * 785.15 K) = (202.6 kPa * V_f * 273.15 K).

V_f = (101.325 kPa * 10.0 L * 785.15 K) / (202.6 kPa * 273.15 K) = 14.37 L

V_f = 14.37 L = 14.4 L

Answer: 14.4 L is the new volume (V_f) of gas.