Question

A balloon at 30.0°C has a volume of 222 mL. If the temperature is increased to 53.1°C and the pressure remains constant, what will the new volume be, in ml? 43 L O of 1 point earned 2 attempts remaining A sample of argon has a volume of 1.2 L at STP. If the temperature is increased to 21°C and the pressure is lowered to 0.80 atm, what will the new volume be, in L? 44 O of 1 point earned 2 attempts remaining A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an elevation where the pressure is 0.72 atm, then the new volume is 1.8 L. What is the temperature in °C) of the air at this elevation? 45 > O of 1 point earned 2 attempts remaining 15.0 L of an ideal gas at 298 K and 3.36 atm are heated to 383 K with a new pressure of 6.30 atm. What is the new volume?

Answer 1

43) Answer :- 392.94 ml

Explanation :-

Given :-

  

We know that at constant pressure, according to Charle's law

  $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ ....(1)

therefore, from equation (1) we have

  $\frac{222 (ml)}{30 (^{0}C)} = \frac{V_{2}}{53.1 (^{0}C)}$

i.e. $V_{2} = \frac{222 (ml) \times 53.1 (^{0}C)}{30 (^{0}C)}$

i.e.   

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44) Answer :- 1.62 L

Explanation :-

Given :- at STP V = 1.2 L

At STP we know , T = 273 K , P = 1 atm

Thus

V1 = 1.2 L , T1 = 273 K , P = 1 atm

V2 = ? , T2 = 21 °C = 294 K , P = 0.8 atm

According to combined gas equation (ideal gas equation) we have

  $\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$ ....(2)

therefore from equation (2) we have

  $\frac{1 (atm) \times 1.2 (L)}{273 (K)} = \frac{0.8 (atm) \times V_{2}}{294 (K)}$

i.e.  

$V_{2} = \frac{1 (atm) \times 1.2 (L) \times 294(K)}{0.8 (atm) \times 273 (K)}$

i.e.

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45) Answer :- 48.84 °C

Explanation :-

Given :-

V1 = 1.2 L , P1 = 1 atm , T1 = 25 °C = 298 K

V2 = 1.8 L , P2 = 0.72 atm , T2 = ?

according to combined gas equation (ideal gas equation) we have

  $\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$.....(3)

therefore from equation (3) we have

$\frac{1 (atm) \times 1.2 (L)}{298 (K)} = \frac{0.72 (atm) \times 1.8 (L)}{T_{2}}$

i.e.  

  $T_{2} = \frac{0.72 (atm) \times 1.8 (L) \times 298(K)}{1(atm) \times 1.2 (L)}$

i.e.   

i.e.

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46) Answer :- 10.3 L

Explanation :-

Given :-

V1 = 15.0 L , P1 = 3.36 atm , T1 = 298 K

V2 = ? , P2 = 6.30 atm , T2 = 383 K

according to combined gas equation (ideal gas equation) we have

  $\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$ ....(4)

i.e.  

$\frac{3.36 (atm) \times 15.0 (L)}{298 (K)} = \frac{6.30 (atm) \times V_{2}}{383 (K)}$

i.e.

  $V_{2} = \frac{3.36 (atm) \times 15.0 (L) \times 383 (K)}{6.30 (atm) \times 298 (K)}$

i.e.

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