States of Matter Test

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States of Matter Test - 58

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Question 1
1 / -0

The vapour pressure of two pure liquids A and B, that form an ideal solution are $$100$$ and $$900$$ torr respectively at temperature T. This liquid solution of A and B is composed of $$1$$ mole of A and $$1$$ mole of B. What will be the pressure, when $$1$$ mole of mixture has been vapourized?

A
$$800$$torr

B
$$500$$torr

C
$$300$$torr

D
None of these

Solution

Vapour pressure of pure A $$P_{A}^{\circ}=100 torr$$
Vapour presuure of pure B $$P_{B}=900 torr$$
$$n_{A}=1 mole$$
$$n_{B}= 1 mole$$
$$x_{A}=\dfrac{1}{1+1}=0.5$$

$$x_B=\dfrac{1}{1+1}=0.5$$

$$P=P_{A}^{\circ}x_{A}+P_{B}^{\circ}x_{B}$$
$$=100\times 0.5+900\times 0.5=500torr$$
Question 2
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A mixture (by weight) of hydrogen and helium is enclosed in a two-litre flask at $$27^o C$$. Assuming ideal kept behaviour, the partial pressure of helium is found to be 0.2atm. Then the concentration of hydrogen would be:

A
0.045

B
8.26

C
0.0162

D
1.62

Solution

$$ \text { Given, } \mathrm{H}_{2} \text { and He gas mixture is enclosed in a flask. } $$

$$ \begin{array}{l} V=2 L \\ T=300 K \end{array} $$
$$ \text { It is given, Partial pressure of He is } 0.2 \mathrm{~atm} \text { . } $$

$$ \text { so, } n_{\text {He }}=\dfrac{P_{\text {He }} V}{R T}=\dfrac{0.2 \times 2}{0.0821 \times 300} \text { moles } $$

$$ =0.01624 \text { moles. } $$
$$ \text { Since, mixture constitutes } 1: 1 \quad H_{2} \text { - He ratio by weight. } $$
$$ \text { Thus, weight of } H_{2}=0.01624 \times 4 \mathrm{gm}=0.06496 \mathrm{gm} $$

$$ \therefore \text { moles of } H_{2}=0.03248 \mathrm{mol} $$

$$ \text { so, concentration of } H_{2}=\dfrac{\text { moles }}{\text { volume }}=\dfrac{0.03248}{2}=0.01624 M $$
Question 3
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If $$0.2\ mol$$ of $$H_{2}(g)$$ and $$2\ mol$$ of $$S(s)$$ are mixed in a $$1\ dm^{3}$$ vessel at $$90^{\circ}C$$, the partial pressure of $$H_{2}S(g)$$ formed according to the reaction, $$H_{2}(g) + S(s)\rightleftharpoons H_{2}S, K_{P} = 6.8\times 10^{-2}$$ would be:

A
$$0.19\ atm$$

B
$$0.38\ atm$$

C
$$0.6\ atm$$

D
$$0.072\ atm$$

Solution

$$H_2 (g) + S(s) \rightleftharpoons H_2S$$
Number mole of $$H_2 \,\ \ n = 0.2\ mol$$
temp $$t = 90^o C + 273 = 363\ K$$
volume = $$1\ dm^3 = 1 litre$$
$$P = \dfrac{nRT}{V} = \dfrac{0.2 \times 0.082 \times 363}{1}$$
$$=5.96\ atm$$
$$H_2 (g) + S(s) \rightleftharpoons H_2 (s)$$
$$5.96 - P_0$$ $$P_0$$
$$K_p = \dfrac{P_0}{5.96 - P_0}$$
$$6.8 \times 10^{-2} = \dfrac{P_0}{5.96 - P_0}$$
$$40.53 \times 10^{-2} - 0.068 P_0 = P_0$$
$$1.068 P_0 = 40.53 \times 10^{-2}$$
$$P_0 = \dfrac{40.53 \times 10^{-2}}{1.068} = 0.3795$$
$$\approx 0.38\ atm$$
Question 4
1 / -0

Two moles of pure liquid 'A' $$(P_{A}^{0} =80mm$$ of Hg) and $$3$$ moles of pure liquid 'B' ($$P_{B}^{0} = 120mm$$ of Hg) are mixed. Assuming ideal behaviour?

A
Vapour pressure of the mixture is $$104$$mm of Hg

B
Mole fraction of liquid 'A' in Vapour pressure is $$0.3077$$

C
Mole fraction of 'B' in Vapour pressure is $$0.692$$

D
Mole fraction of 'B' in Vapour pressure is $$0.785$$

Solution

$$P_{A}^{\circ}=80 mm Hg$$ $$P_{B}^{\circ}=120 mm Hg$$
$$n_{A}=2 moles$$ $$n_{B}=3 moles$$
$$x_{A}=\dfrac{2}{2+3}=\dfrac{2}{5}=0.4$$

$$x_{B}=\dfrac{3}{5}=0.6$$
$$P_{A}=P_{A}^{\circ}x_{A}=80\times 0.4=32 mm Hg$$
$$P_{B}=P_{B}^{\circ}x_{B}=120\times 0.6=72 mm Hg$$
Total vapour pressure $$P=P_{A}+P_{B}$$
$$=32+72=104 mm Hg$$
Question 5
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The virial equation for 1 mole of a real gas is written as:
PV=RT

[1+$$\frac{A}{V}+\frac{B}{V^2}+\frac{C}{V^3}+$$................to the higher power of n]

Where A, B and C are known as virial coefficients. If Vander Waal's equation is written in virial form, then what will be the value of B?

A
$$a-\cfrac{b}{RT}$$

B
$$b^3$$

C
$$b-\cfrac{a}{RT} $$

D
$$b^2$$

Solution

A virial equation is an equation of state of gases that has additional terms beyond that for an ideal gas, which account for the interactions between the molecules.

The equation of state for a van der Waals gas is:

$$\boxed{\left(p+\dfrac{a}{V^2_m}\right)(V_m-b)=RT}$$

This leads to the following virial expansion:
$$\dfrac{pV_m}{RT}+1+\dfrac{B}{V_m}+\dfrac{C}{V^2_m}+-----(1)$$

The van der Waals equations of state can be rewritten as:

$$pV=\dfrac{RT}{V-b}+\dfrac{a}{V^2}=\dfrac{RT}{V}\left(1-\dfrac{b}{V}\right)^{-1}+\dfrac{a}{V^2}$$

and using the binomial expansion, the terms in brackets can be expanded into a series, resulting in
$$\dfrac{pV}{RT}=1+\dfrac{1}{V}(b-\dfrac{a}{RT})+\left(\dfrac{b}{V}\right)^2+\left(\dfrac{b}{V}\right)^3+----,$$

which is in the same form as the virial expansion is equation (1) with
$$B(T)=b-\dfrac{a}{RT}$$

If the Van der Waal's equation is written in the virial form then the value of $$B=b-\dfrac{a}{RT}$$.

Hence, option C is correct.
Question 6
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Vapour pressure of $$CCl_4$$ at $$25^0$$C is $$143$$mm Hg. $$0.5$$ gm of a non-volatile solute (mol.wt.$$65$$) is dissolved in $$100$$ ml of $$CCl_4$$. Find the vapour pressure of the solution. (Density of CCl_4 = $$1.58 gm/cm^3$$)

A
$$141.93$$mm

B
$$94.39$$mm

C
$$199.34$$mm

D
$$143.99$$mm

Solution
Question 7
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Which of the following relationships between partial pressure, volume and temperature is correct?
(i) $$P = \dfrac {nRT}{V}$$
(ii) $$P_{total} = p_{1} + p_{2} + p_{3}$$
(iii) $$P_{total} = (n_{1} + n_{2} + n_{3}) \dfrac {RT}{V}$$.

A
(i) and (ii)

B
(i) and (iii)

C
(ii) and (iii)

D
(i), (ii) and (iii)

Solution

$$i.$$ Ideal gas equation $$\Longrightarrow PV=nRT\Longrightarrow P=\cfrac{nRT}{V}$$
$$ii.$$ Dalton's law of partial pressure, $$\Longrightarrow P_{total}=P_1+P_2+P_3$$
$$iii.$$ Partial pressure , $$P_1,P_2,P_3\dots P_K$$
According to ideal gas equation, $$P=\cfrac{nRT}{V}\\ \Longrightarrow P_1=n_1(\cfrac{RT}{V})$$
$$P_{total}=(n_1+n_2+n_3)(\cfrac{RT}{V})$$
Question 8
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A rigid and insulated tank of 3 $$m^3$$ volume is divided into two components. One compartment of volume of 2 $$m^3$$ contains an ideal gas at 0.8314 MPa and 400 K and while the second compartment of volume 1 $$m^3$$ contains the same gas 8.314 MPa and 500 k. If the partition between the two compartments is ruptured, the final temperature of the gas is:

A
420 K

B
450 K

C
480 K

D
None of these

Solution

First compartment
$$P_1=0.8314\ M\ Pa\ V_1=2\ xm^2\ T_1=400\ K$$
Second compartment
$$P_2=8.314\ M\ Pa\ V_2=1\ m^3\ T_2=500\ K$$
$$n_1=\dfrac {P_1 V_1}{RT_1}=\dfrac {0.8314\times 10^6 \times 2}{8.314 \times 400}=500$$
$$n_2=\dfrac {P_2 V_2}{RT_2}=\dfrac {8.314\times 1\times 10^6 }{8.314 \times 500}=200$$
After Ruptewe of compartment
$$n=500+200=700\ V=2\ P_1=3\ m^3$$
$$P=0.8314\ M\ Pq$$
$$T=\dfrac {PV}{nR}=\dfrac {3\times 8.314\times 10^6}{700\times 8.314}$$
$$=\dfrac {3000}{7}$$
$$\simeq 420\ K$$
Question 9
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The pressure of a mixture of equal weights of two gases $$X$$ and $$Y$$ with molecular weight $$4$$ and $$40$$ respectively is $$1.1\ atm$$. The partial pressure of the gas $$X$$ in the mixture is :

A
$$1\ atm$$

B
$$0.1\ atm$$

C
$$0.15\ atm$$

D
$$0.5\ atm$$

Solution

Let $$w\ g$$ of each gas is taken.
Mole fraction of $$X = \dfrac {\dfrac {w}{4}}{\dfrac {w}{4} + \dfrac {w}{40}} = \dfrac {10}{11}$$
Partial pressure of $$X = P_{total}\times Mole\ fraction$$
$$= 1.1\times \dfrac {10}{11} = 1\ atm$$.
Question 10
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An open flask contains air at $$27^{\circ}C$$. At what temperature should it be heated so that $$1/3rd$$ of air present in it goes out?

A
$$177^{\circ}C$$

B
$$100^{\circ}C$$

C
$$300^{\circ}C$$

D
$$150^{\circ}C$$

Solution

Let initial number of moles of air at $$27^{\circ}C (300\ K) = n$$
At temperature $$T\ K$$, the no. of moles left $$= n - \dfrac {n}{3} = \dfrac {2n}{3}$$
At constant pressure and volume, $$n_{1}T_{2} = n_{2}T_{2}$$
$$n\times 300 = \dfrac {2n}{3}\times T\Rightarrow T = 450\ K$$ or $$177^{\circ}C$$.

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States of Matter Test - 58

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States of Matter Test - 58

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Question 1

$$800$$torr

$$500$$torr

$$300$$torr

None of these

Solution

Question 2

0.045

8.26

0.0162

1.62

Solution

Question 3

$$0.19\ atm$$

$$0.38\ atm$$

$$0.6\ atm$$

$$0.072\ atm$$

Solution

Question 4

Vapour pressure of the mixture is $$104$$mm of Hg

Mole fraction of liquid 'A' in Vapour pressure is $$0.3077$$

Mole fraction of 'B' in Vapour pressure is $$0.692$$

Mole fraction of 'B' in Vapour pressure is $$0.785$$

Solution

Question 5

$$a-\cfrac{b}{RT}$$

$$b^3$$

$$b-\cfrac{a}{RT} $$

$$b^2$$

Solution

Question 6

$$141.93$$mm

$$94.39$$mm

$$199.34$$mm

$$143.99$$mm

Solution

Question 7

(i) and (ii)

(i) and (iii)

(ii) and (iii)

(i), (ii) and (iii)

Solution

Question 8

420 K

450 K

480 K

None of these

Solution

Question 9

$$1\ atm$$

$$0.1\ atm$$

$$0.15\ atm$$

$$0.5\ atm$$

Solution

Question 10

$$177^{\circ}C$$

$$100^{\circ}C$$

$$300^{\circ}C$$

$$150^{\circ}C$$

Solution

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