Question 1

Multiple choiceMedium1 mark

A fixed amount of an ideal gas has pressure 8.0×104 Pa8.0 \times 10^4\ \mathrm{Pa} volume 6.0×10−3 m36.0 \times 10^{-3}\ \mathrm{m^3} and temperature 300 K300\ \mathrm{K} It is changed to volume 3.0×10−3 m33.0 \times 10^{-3}\ \mathrm{m^3} and temperature 450 K450\ \mathrm{K} What is its final pressure?

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Answer: D. For a fixed amount of gas, P∝T/VP \propto T/V Thus, P2=P1(V1/V2)(T2/T1)=(8.0×104)(2.0)(1.5)=2.4×105 PaP_2=P_1(V_1/V_2)(T_2/T_1)=(8.0\times10^4)(2.0)(1.5)=2.4\times10^5\ \mathrm{Pa}

A: reverses the temperature ratio, using T1/T2T_1/T_2 instead of T2/T1T_2/T_1
B: accounts for the temperature increase but omits the volume decrease.
C: accounts for the volume decrease but omits the temperature increase.
Remember: at fixed amount, pressure increases with temperature and decreases with volume.

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