In a process the density of a gas remains constant. If the temperature is doubled, then the change in the pressure will be :
Given that density \(\rho\) is constant, we use the ideal gas law:
\[
\frac{P}{T} = \text{constant (since } \rho \text{ is constant)}
\]
If the temperature \(T\) is doubled, then:
\[
\frac{P_2}{T_2} = \frac{P_1}{T_1} \Rightarrow \frac{P_2}{2T_1} = \frac{P_1}{T_1}
\]
Thus, \(P_2 = 2P_1\), meaning the pressure also doubles, resulting in a 100% increase.
‘A’ is a closed vessel of volume V and contains O2 at pressure P and temperature T. ‘B’ is another closed vessel of same volume and it contains H2 at same temperature and 2P pressure. Ratio of masses of O2 and H2 in vessel ‘A’ and ‘B’ is :
Using the ideal gas law \( PV = nRT \), we find the moles \( n \) in each vessel:
1. For vessel \( A \) with \( O_2 \):
\[
n_{\text{O}_2} = \frac{PV}{RT}
\]
2. For vessel \( B \) with \( H_2 \) at \( 2P \):
\[
n_{\text{H}_2} = \frac{2PV}{RT} = 2 \times \frac{PV}{RT}
\]
Now, the mass \( m = n \times \text{molar mass} \):
- Mass of \( O_2 \) in \( A = n_{\text{O}_2} \times 32 = \frac{PV}{RT} \times 32 \)
- Mass of \( H_2 \) in \( B = n_{\text{H}_2} \times 2 = 2 \times \frac{PV}{RT} \times 2 \)
So, the mass ratio is:
\[
\frac{\text{mass of } O_2}{\text{mass of } H_2} = \frac{\frac{PV}{RT} \times 32}{2 \times \frac{PV}{RT} \times 2} = \frac{32}{4} = 8:1
\]
One mole of an ideal gas undergoes a process \( P=\frac{_{P_{0}}}{1+\left( \frac{V_{0}}{V} \right)^{2}} \) , Here P0 and V0 are constants. Change in temperature of the gas when volume is changed from V = V0 to V = 2V0 is :
For one mole of an ideal gas, we use the ideal gas law:
\[
PV = RT
\]
Given \( P = \frac{P_0}{1 + \left( \frac{V_0}{V} \right)^2} \), find \( T \) at \( V = V_0 \) and \( V = 2V_0 \).
1. When \( V = V_0 \):
\[
P = \frac{P_0}{1 + \left( \frac{V_0}{V_0} \right)^2} = \frac{P_0}{1 + 1} = \frac{P_0}{2}
\]
\[
T_1 = \frac{PV}{R} = \frac{\frac{P_0}{2} \cdot V_0}{R} = \frac{P_0 V_0}{2R}
\]
2. When \( V = 2V_0 \):
\[
P = \frac{P_0}{1 + \left( \frac{V_0}{2V_0} \right)^2} = \frac{P_0}{1 + \frac{1}{4}} = \frac{P_0}{\frac{5}{4}} = \frac{4P_0}{5}
\]
\[
T_2 = \frac{PV}{R} = \frac{\frac{4P_0}{5} \cdot 2V_0}{R} = \frac{8P_0 V_0}{5R}
\]
3. Change in temperature:
\[
\Delta T = T_2 - T_1 = \frac{8P_0 V_0}{5R} - \frac{P_0 V_0}{2R} = \frac{11P_0 V_0}{10R}
\]
In the diagram as shown, find parameters representing x and y axis and also parameter z, if z1 = z2, z3 = z4 and z2 > z3 :
No solution provided for this question.
Two gases A and B having the same temperature T, same pressure P and same volume V are mixed. If the mixture is at the same temperature T and occupies a volume V, the pressure of the mixture is :
Since both gases \( A \) and \( B \) have the same temperature \( T \), pressure \( P \), and volume \( V \), each gas has the same number of moles, \( n \), by the ideal gas law:
\[
PV = nRT \Rightarrow n = \frac{PV}{RT}
\]
When gases \( A \) and \( B \) are mixed, the total number of moles becomes \( 2n \). Since temperature \( T \) and volume \( V \) remain the same, the total pressure \( P_{\text{mixture}} \) will be:
\[
P_{\text{mixture}} V = (2n)RT
\]
\[
P_{\text{mixture}} = 2P
\]
Answer: The pressure of the mixture is \( 2P \).
In the gas equation PV = nRT, the value of universal gas constant would depend only on :
In the gas equation \( PV = nRT \), the universal gas constant \( R \) has a fixed value but its numerical value depends on the units of \( P \), \( V \), and \( T \).
Answer: The value of the universal gas constant depends only on the units of measurement.
The quantity (PV/kT) represents :
The expression \(\frac{PV}{kT}\) can be analyzed using the ideal gas law:
\[
PV = NkT
\]
where:
- \( P \) is pressure,
- \( V \) is volume,
- \( N \) is the number of molecules,
- \( k \) is Boltzmann's constant, and
- \( T \) is temperature.
Rearranging, we get:
\[
\frac{PV}{kT} = N
\]
Answer: This quantity represents the "number of molecules in the gas".
When the temperature of a gas in a metal box is increased from 27°C to 87°C, the initial pressure of 2 atmosphere changes to :
Given:
- Initial temperature, \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \)
- Final temperature, \( T_2 = 87^\circ \text{C} = 360 \, \text{K} \)
- Initial pressure, \( P_1 = 2 \, \text{atm} \)
Using the pressure-temperature relationship for a gas at constant volume:
\[
\frac{P_2}{P_1} = \frac{T_2}{T_1}
\]
Solving for \( P_2 \):
\[
P_2 = P_1 \times \frac{T_2}{T_1} = 2 \times \frac{360}{300} = 2 \times 1.2 = 2.4 \, \text{atm}
\]
Answer: \( P_2 = 2.4 \, \text{atm} \)
A large flask contains air at 27°C. In order to expel half the mass of air from the flask, the flask should be heated to :
Given:
- Initial temperature, \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \)
- We want to expel half the mass of air, meaning the final mass, \( m_2 = \frac{m_1}{2} \).
Using the ideal gas law, \( PV = nRT \), and since pressure and volume are constant, \( \frac{m}{T} = \text{constant} \).
Thus,
\[
\frac{m_1}{T_1} = \frac{m_2}{T_2}
\]
Substitute \( m_2 = \frac{m_1}{2} \):
\[
\frac{m_1}{300} = \frac{\frac{m_1}{2}}{T_2}
\]
Solving for \( T_2 \):
\[
T_2 = 2 \times 300 = 600 \, \text{K}
\]
Converting back to Celsius:
\[
T_2 = 600 - 273 = 327^\circ \text{C}
\]
Answer: \( T_2 = 327^\circ \text{C} \)
An ideal gas contained in a cylinder undergoes a thermodynamic process during which pressure relates to volume as \( P=\frac{A}{1+\left( \frac{B}{V} \right)^{2}}\), where A and B are constants. As the volume of the gas is changed from V = B to V = 2B, its change of temperature can be expressed as :
Given:
\[
P = \frac{A}{1 + \left( \frac{B}{V} \right)^2}
\]
Using the ideal gas equation, \( PV = nRT \), for initial and final states, we can express the temperature change.
Step 1: Initial State (at \( V = B \))
\[
P_1 = \frac{A}{1 + \left( \frac{B}{B} \right)^2} = \frac{A}{2}
\]
\[
T_1 = \frac{P_1 V}{R} = \frac{\left(\frac{A}{2}\right) B}{R} = \frac{AB}{2R}
\]
Step 2: Final State (at \( V = 2B \))
\[
P_2 = \frac{A}{1 + \left( \frac{B}{2B} \right)^2} = \frac{A}{1 + \frac{1}{4}} = \frac{A}{\frac{5}{4}} = \frac{4A}{5}
\]
\[
T_2 = \frac{P_2 V}{R} = \frac{\left(\frac{4A}{5}\right) (2B)}{R} = \frac{8AB}{5R}
\]
Step 3: Temperature Change
\[
\Delta T = T_2 - T_1 = \frac{8AB}{5R} - \frac{AB}{2R} = \frac{16AB - 5AB}{10R} = \frac{11AB}{10R}
\]
Answer: \(\frac{11AB}{10R}\)