Home
Class 12
PHYSICS
Two containers of equal volume contain t...

Two containers of equal volume contain the same gas at pressure `P_(1)` and `P_(2)` and absolute temperature `T_(1)` and `T_(2)`, respectively. On joining the vessels, the gas reaches a common pressure `P` and common temperature `T`. The ratio `P//T` is equal to

A

`P_1/T_1+P_2/T_2`

B

`(P_1T_1+P_2T_2)/((T_1+T_2))`

C

`(P_1T_2+P_2T_1)/((T_1+T_2))`

D

`P_1/(2T_1)+P_2/(2T_2)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use the ideal gas law and the concept of conservation of moles when two containers are combined. Let's break down the solution step by step. ### Step 1: Write the Ideal Gas Law for Each Container For the first container, we can express the number of moles (μ₁) using the ideal gas law: \[ \mu_1 = \frac{P_1 V}{RT_1} \] Where: - \(P_1\) = Pressure in the first container - \(V\) = Volume of the container (equal for both) - \(R\) = Universal gas constant - \(T_1\) = Absolute temperature in the first container For the second container, the number of moles (μ₂) is given by: \[ \mu_2 = \frac{P_2 V}{RT_2} \] ### Step 2: Total Moles After Joining the Containers When the two containers are joined, the total number of moles of gas remains the same: \[ \mu = \mu_1 + \mu_2 \] Substituting the expressions from Step 1: \[ \mu = \frac{P_1 V}{RT_1} + \frac{P_2 V}{RT_2} \] ### Step 3: Express the Common Pressure and Temperature After joining, let the common pressure and temperature be \(P\) and \(T\), respectively. The total number of moles can also be expressed as: \[ \mu = \frac{P (2V)}{RT} \] Since the volume of both containers is equal, the total volume is \(2V\). ### Step 4: Set the Two Expressions for Moles Equal Equating the two expressions for the number of moles: \[ \frac{P (2V)}{RT} = \frac{P_1 V}{RT_1} + \frac{P_2 V}{RT_2} \] We can cancel \(V\) and \(R\) from both sides: \[ \frac{2P}{T} = \frac{P_1}{T_1} + \frac{P_2}{T_2} \] ### Step 5: Solve for the Ratio \(\frac{P}{T}\) Rearranging the equation gives us: \[ \frac{P}{T} = \frac{1}{2} \left( \frac{P_1}{T_1} + \frac{P_2}{T_2} \right) \] ### Final Result Thus, the ratio \(\frac{P}{T}\) is: \[ \frac{P}{T} = \frac{P_1}{2T_1} + \frac{P_2}{2T_2} \] ### Conclusion The correct answer is: \[ \frac{P}{T} = \frac{P_1}{2T_1} + \frac{P_2}{2T_2} \]
Promotional Banner

Similar Questions

Explore conceptually related problems

Two vessels of volume (V_1) and (V_2) contain the same ideal gas. The pressures in the vessels are (P_1) and (P_2) and the temperatures are (T_1) and (T_2) respectively . The two vessels are now connected to each other through a narrow tube. Assuming that no heat is exchanged between the surrounding and the vessels, find the common pressure and temperature attained after the connection.

Two idential container joined by a small pipe initially contain the same gas at pressure p_(0) and absolute temperature T_(0) . One container is now maintained at the same temperature while the other is heated to 2T_(0) . The common pressure of the gas

An ideal gas has pressure P, volume V and temperature T. Select the correct option.

An ideal gas in a thermally insulated vessel at internal pressure =P_(1) , volume =V_(1) and absolute temperature =T_(1) expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are P_(2), V_(2) and T_(2) respectively. For this expansion.

An ideal gas in a thermally insulated vessel at internal pressure =P_(1) , volume =V_(1) and absolute temperature =T_(1) expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are P_(2), V_(2) and T_(2) respectively. For this expansion.

An ideal gas in a thermally insulated vessel at internal pressure = P_(1) , volume = V_(1) and absolute temperature = T_(1) expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are P_(2), V_(2) and T_(2) , respectively. For this expansion.

Two identical containers joned by a small pipe initially contain the same gas at pressue p_(o) and abosolute temperature T_(o^.) One container is now mantained at the same temperature while the other is heated to 2T_(0^.) The commmon pressure of the gases will be

Two adiabatic containers have volumes V_(1) and V_(2) respectively. The first container has monoatomic gas at pressure p_(1) and temperature T_(1) . The second container has another monoatomic gas at pressure p_(2) and temperature T_(2) . When the two containers are connected by a narrow tube, the final temperature and pressure of the gases in the containers are P and T respectively. Then

Two cylinder having m_(1)g and m_(2)g of a gas at pressure P_(1) and P_(2) respectively are put in cummunication with each other, temperature remaining constant. The common pressure reached will be

A sample of an ideal gas occupies a volume V at pressure P and absolute temperature T. The masss of each molecule is m, then the density of the gas is