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Thermal Properties of Matter Test - 37

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Thermal Properties of Matter Test - 37
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  • Question 1
    1 / -0
    A sample of an ideal gas occupies a volume of 'V' at a pressure 'P' and absolute temperature 'T'. The mass of each molecule is 'm'. The equation for density is 
    Solution
    We know that PV = nRT =(m/M)RT
    Where M = Molecular weight
    Now $$P\times \left(\displaystyle \frac{m}{d}\right)$$ = $$\left(\displaystyle\ \frac{m}{M} \right)RT$$
    where d = density of gas
    $$\displaystyle\ \frac{p}{d}$$ = $$\displaystyle\ \frac{RT}{M}$$ = $$\displaystyle\ \frac{kN_{A}T}{M}$$
    Where $$R$$ = $$kN_{A}$$, k is Boltzmann constant
    But $$\displaystyle\ \frac{ M}{N_{A}}$$ = m = mass of each molecule so
    So, $$d$$ = $$\displaystyle\ \frac{P\times m}{kT}$$
  • Question 2
    1 / -0
    An electric heater kept in vacuum is heated continuously by passing an electric current. Its temperature
    Solution
    Unless the heat generation rate is matched by the rate of radiation, temperature will continue to rise.
    After that till the heat rate remains constant, temperature remains constant.
  • Question 3
    1 / -0
    A liquid used as a coolant in automobile engine is 
    Solution
    Radiators are heat exchangers used for cooling internal combustion engines, mainly in automobiles.
    Internal combustion engines are often cooled  by circulating a liquid called engine coolant through the engine block, where it is heated, then through a radiator where it loses heat to the atmosphere, and then returned to the engine. Engine coolant is usually water-based.

  • Question 4
    1 / -0
    The air density at Mount Everest is less than that at the sea level. It is found by mountaineers that for one trip lasting a few hours, the extra oxygen needed by them corresponds to 30,000cc at sea level (pressure 1 atmosphere, temperature 27$$^o$$C). Assume that the temperature around Mount Everest is -$$73^o$$C and that the oxygen cylinder has a capacity of 5.2 litters. The pressure at which oxygen be filled (at site) in the cylinder is
    Solution
    Since moles of gas must remain constant,

    $$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } $$

    $$\dfrac { 1(30000) }{ 300 } =\dfrac { P(5200) }{ 143 } $$

    $$P=3.86atm$$

  • Question 5
    1 / -0
    A vessel contains air at a temperature of $$15^{0}C$$ and 60% R.H. What will be the R.H if it is heated to $$20^{0}C$$? (S.V.P at $$15^{0}C$$ is 12.67 & at $$20^{0}C$$ is 17.36mm of Hg respectively)
    Solution
    By definition, 
    $$\displaystyle\ \frac{V.P  }{S.V.P  }\times100$$ =$$\displaystyle\ \frac{60}{100}$$ (both VP and SVP are in room temperature)
    $$= \displaystyle\ \frac{(V.P)_{15}}{12.67}$$
    $$(V.P)_{15} = 7.6 mm$$ $$of$$ $$Hg$$
    Now since unsaturated vapour obeys gas equation & mass of water remains constant so 
    $$P$$ = $$\left (\displaystyle\ \frac{nRT}{V} \right)$$ 
    $$\Rightarrow$$ $$P\ \alpha \ T$$
    $$\displaystyle\ \frac{(V.P)_{15}}{(V.P)_20}$$
    $$= \displaystyle\ \frac{273+15}{273+20}$$
    $$\Rightarrow$$ $$(RH)_{20}$$= 7.73 mm of Hg
    So $$(R.H)_{20} =\displaystyle\ \frac{(V.P)_{20}}{(S.V.P)_{20}}\times 100$$
    $$= 44.5$$%
  • Question 6
    1 / -0

    Directions For Questions

    Consider air to be a diatomic gas with average mole moasses 29g/mole. A mass 1.45kg of air is contained in a cylinder with a piston at $$27^{0}C$$ and pressure $$1.5\times10^{5} N/m^{2}$$. Energy is given to the system as heat and the system is allowed to expand till the final pressure $$3.5\times10^{5} N/m^{2}$$. As the gas expands, pressure and volume follow the relation
    $$ V= KP^{2}$$ where K is constant (R =8.3J/(mole-K)

    ...view full instructions

    Final volume of the system will be nearly -
    Solution
    $$PV = nRT$$
    Initial volume 
    $$= \displaystyle\ \frac{ 1.45\times10^{3}}{29} \times \displaystyle\ \frac{8.3\times300}{1.5\times10^{5}}m^{3}$$
    $$\displaystyle\ \frac{V_{i}}{P_{i^{2}}}$$ =$$\displaystyle\ \frac{V_{f}}{P_{f}^{2}}$$ $$\Rightarrow$$ $$V_{f}$$ = $$\left (\displaystyle\ \frac{3.5}{1.5}\right)^{2} V_{i}$$ = $$4.5 m^{3}$$
    $$T = \displaystyle\ \frac{P_{f}V_{f}}{nR} = 3800K$$
    $$\triangle U = nC_{V}\triangle T = n \displaystyle\ \frac{5R}{2} \triangle T = 3.6\times10^{6}J$$
  • Question 7
    1 / -0
    A vessel has 6g of hydrogen at pressure P and temperature 500K. A small hole is made in it so that hydrogen leaks out. How much hydrogen leaks out if the final pressure is $$P/2$$ and temperature falls to 300K?
    Solution
    $$PV$$ = $$\displaystyle\ \frac{m}{M}RT$$
    Initally, $$PV$$ = $$\displaystyle\ \frac{6}{M}R\times500$$
    Finally, $$\displaystyle\ \frac{P}{2}V$$ = $$\displaystyle\ \frac{(6-x)}{M}R\times300$$ ( if x g gas leaks out)
    Hence, $$2 = \displaystyle\ \frac{6}{6-x}\times\displaystyle\ \frac{5}{3}$$
    $$\therefore x = 1$$gram
  • Question 8
    1 / -0
    A graph is plotted with PV/T on y-axis and mass of the gas along x-axis for different gases. The graph is 
    Solution
    $$\displaystyle\ \frac{PV}{T}$$ = nR = $$\left( \displaystyle\ \frac{m}{M}\right)R$$ 
    Or, $$\displaystyle\ \frac{PV}{T} = \left (\displaystyle\ \frac{R}{M}\right)m$$
    i.e., $$\displaystyle\ \frac{PV}{T}$$ versus on graph is straight line passing through origin with slope R/M. i.e., the slope depends on molecular mass of the gas M and is different for different gases
  • Question 9
    1 / -0
    If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by $$1^{0}C$$, the initial temperature must be 
    Solution
    We know that $$\displaystyle\ \frac{P_{1}}{T_{1}}$$ = constant for constant volume.
    Let the initial pressure be $$P_1$$ and initial temperature be $$T_1$$.
    Thus final pressure $$P_2 =1.004 P_1 $$ and final temperature  $$T_2 = T_1 + 1$$.
    Using  $$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$$
    So, $$\displaystyle\ \frac{P_{1}}{T_{1}}$$ = $$\displaystyle\ \frac{1.004 P_{1}}{T_{1}+1}$$
    Or  $$T_1 + 1 = 1.004 T_1$$ 
    Or  $$0.004 T_1 = 1$$
    $$\Rightarrow  T_1 =250K$$
  • Question 10
    1 / -0
    An ideal gas is found to obey an additional law $$VP^{2}$$ = constant. The gas is initially at temperature T and volume V. When it expands to a volume 2V, the temperature becomes :
    Solution
    We are given an ideal gas which is following an additional law,
    $$PV^2=$$ a constant            ....................(1)

    As the gas is ideal, the gas law is
    $$PV=nRT$$

    Taking the value of $$P$$ from the above equation, we get
    $$P=\dfrac{nRT}{V}$$

    Putting this value in equation (1), we get
    $$\dfrac{nRT(V)^2}{V} =$$ constant
    As $$n$$ and $$R$$ are constant, above equation becomes,
    $$TV=$$ a constant                .....................(2)

    Now, the gas is expanding from initial volume $$V_1$$ to final volume $$V_2$$ by $$2$$ units, it becomes
    $$V_2=2V_1$$

    From equation (2), we get
    $$T_1 V_1 = T_2 V_2$$
    $$T_1 V_1 = T_2 (2V_1)$$
    $$T_2=\dfrac{T_1}{2}$$
    So the correct answer is $$\dfrac{T}{2}$$
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